Refraction at Spherical Surfaces and by Lenses

Ray Optics and Optical Instruments · CBSE

Theory

What this subtopic is examined on

The derivation of $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2-n_1}{R}$ for refraction at a single spherical surface, with a ray diagram (set repeatedly as 5 marks).
The derivation of the lens maker's formula from it, and of the thin lens formula from the lens maker's formula.
Power of a lens and its SI unit; combinations of thin lenses in contact and the derivation $\dfrac1f = \dfrac1{f_1}+\dfrac1{f_2}$ .
How a lens behaves in a liquid, when cut, and when the wavelength changes.
Numericals: lens maker's formula, lens formula, virtual objects, lens–mirror and lens–lens combinations, displacement method, liquid-lens experiment.

Refraction at a single spherical surface

Diagram in words

A convex spherical surface separates a rarer medium of index $n_1$ (left) from a denser medium of index $n_2$ (right). Mark the pole $P$ , the centre of curvature $C$ to the right at distance $R$ , and the principal axis through them.
A point object $O$ lies on the axis in the rarer medium. Draw two rays from $O$ : one along the axis, striking the surface normally and going straight on; the other, $ON$ , striking the surface at a point $N$ near the pole. Join $CN$ — this is the normal at $N$ . The ray refracts towards the normal and crosses the axis at $I$ , the real image, in the denser medium.
Let $\alpha$ , $\beta$ , $\gamma$ be the small angles that $ON$ , $IN$ and $CN$ make with the axis, and drop the perpendicular $NM$ from $N$ onto the axis.

Derivation

In triangle $ONC$ , the angle of incidence $i$ is the exterior angle, so $i = \alpha + \gamma$ .
In triangle $NCI$ , $\gamma$ is the exterior angle, so $\gamma = r + \beta$ , i.e. $r = \gamma - \beta$ .
Snell's law for small angles: $n_1\sin i = n_2\sin r$ becomes $n_1 i = n_2 r$ , so

$n_1(\alpha+\gamma) = n_2(\gamma-\beta) \qquad\Rightarrow\qquad n_1\alpha + n_2\beta = (n_2-n_1)\gamma$

For paraxial rays, $M$ lies very close to $P$ and the angles equal their tangents:

$\alpha \approx \dfrac{MN}{OM}, \qquad \beta \approx \dfrac{MN}{MI}, \qquad \gamma \approx \dfrac{MN}{MC}$

Substituting and cancelling $MN$ :

$\dfrac{n_1}{OM} + \dfrac{n_2}{MI} = \dfrac{n_2-n_1}{MC}$

Apply the Cartesian sign convention: $OM = -u$ , $MI = +v$ , $MC = +R$ . Then

$\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2-n_1}{R}$

This holds for any spherical refracting surface, convex or concave, real or virtual image. Remember: the index of the medium the light enters goes with $v$ .
Worked example: a point source in air 100 cm from a glass surface ( $n = 1.5$ , $R = +20$ cm). With $u = -100$ cm, $\dfrac{1.5}{v} + \dfrac{1}{100} = \dfrac{0.5}{20}$ , giving $\dfrac{1.5}{v} = \dfrac{3}{200}$ and $v = +100$ cm — a real image 100 cm inside the glass.
A virtual-image case: a point object in air at $u = -R$ from a convex glass surface of radius $R$ and $\mu_g = 1.5$ . Then $\dfrac{1.5}{v} + \dfrac1R = \dfrac{0.5}{R}$ , so $\dfrac{1.5}{v} = -\dfrac{0.5}{R}$ and $v = -3R$ : the image lies on the same side as the object, in the air, and is virtual.

Refraction through a solid sphere

A sphere has two spherical surfaces, so apply the relation twice: at the near surface with $R_1 = +r$ , then at the far surface with $R_2 = -r$ , taking the first image as the object for the second and measuring its distance from the far pole (a distance $2r$ further along the axis). An image that would form beyond the far surface is a virtual object there, with $u$ positive.
The chord geometry shortcut. Because the two radii drawn to the entry and exit points make an isosceles triangle with the chord inside the glass, the angle of incidence inside at the exit face equals the angle of refraction at the entry face. So $e = i$ always: a ray entering a sphere emerges parallel to no fixed line in general, but its angle with the local normal repeats.
Standard question: a ray strikes a sphere of $n = \sqrt3$ at $60^\circ$ to the diameter $AB$ and emerges parallel to $AB$ . At entry, $\sin r = \dfrac{\sin 60^\circ}{\sqrt3} = \dfrac12$ , so $r = 30^\circ$ ; by the symmetry above the angle of incidence at the exit face is also $30^\circ$ , and $\sin e = \sqrt3\sin 30^\circ = \dfrac{\sqrt3}{2}$ , so the angle of emergence is $60^\circ$ .

Lens maker's formula

Diagram in words

A thin double-convex lens of material index $n_2$ in a medium of index $n_1$ , with the first surface (radius $R_1$ , pole $B$ ) meeting the light first and the second surface (radius $R_2$ , pole $D$ ) second. The object $O$ is on the axis to the left.
Show refraction at the first surface alone forming an intermediate image $I_1$ inside (or beyond) the lens, and then refraction at the second surface taking $I_1$ as its object and forming the final image $I$ .

Derivation

At the first surface, light goes from $n_1$ into $n_2$ , object at $u$ , intermediate image at $v'$ :

$\dfrac{n_2}{v'} - \dfrac{n_1}{u} = \dfrac{n_2-n_1}{R_1} \qquad \text{...(i)}$

At the second surface, $I_1$ acts as the object and light goes from $n_2$ back into $n_1$ , final image at $v$ :

$\dfrac{n_1}{v} - \dfrac{n_2}{v'} = \dfrac{n_1-n_2}{R_2} \qquad \text{...(ii)}$

The lens is thin, so the two poles are effectively the same point and $v'$ is common. Adding (i) and (ii), the terms in $v'$ cancel:

$\dfrac{n_1}{v} - \dfrac{n_1}{u} = (n_2-n_1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

Dividing by $n_1$ and writing $n_{21} = n_2/n_1$ :

$\dfrac{1}{v} - \dfrac{1}{u} = (n_{21}-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) \qquad \text{...(iii)}$

Put the object at infinity, $u\to-\infty$ ; then the image forms at the focus, $v = f$ :

$\dfrac{1}{f} = (n_{21}-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

This is the lens maker's formula. It is true for a concave lens as well: there $R_1$ is negative and $R_2$ positive, so $f$ comes out negative.

Thin lens formula

Comparing (iii) with the lens maker's formula, the right-hand sides are identical, so

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

Derived here for a real image formed by a convex lens, it is valid for convex and concave lenses and for real and virtual images alike.
The two foci $F$ and $F'$ of a thin lens are equidistant from the optical centre. The focus on the side of the original source is the first focal point; the other is the second focal point.

Ray-diagram rules for lenses

A ray parallel to the principal axis passes, after refraction, through the second principal focus $F'$ of a convex lens, or appears to diverge from the first focus $F$ of a concave lens.
A ray through the optical centre passes undeviated.
A ray through the first principal focus of a convex lens emerges parallel to the axis; a ray directed at the second focus of a concave lens emerges parallel to the axis.

Magnification

$m = \dfrac{h'}{h} = \dfrac{v}{u}$

No minus sign here, unlike the mirror. $m$ positive means an erect, virtual image; $m$ negative means an inverted, real image.
Conditions asked for by CBSE: with the object beyond the focus of a converging lens the image is real and inverted and $m$ is negative; with the object between the focus and the optical centre the image is virtual, erect and enlarged and $m$ is positive.

Object (convex lens)	Image	Nature	Size
At infinity	At $F'$	Real, inverted	Point
Beyond $2F$	Between $F'$ and $2F'$	Real, inverted	Diminished
At $2F$	At $2F'$	Real, inverted	Same size
Between $F$ and $2F$	Beyond $2F'$	Real, inverted	Magnified
At $F$	At infinity	Real, inverted	Very large
Between $F$ and lens	Same side as object	Virtual, erect	Magnified

A concave lens always gives a virtual, erect, diminished image between the optical centre and $F$ , for any real object.
As an object approaches the focus of a converging lens from far away, the real image moves away from the lens, faster and faster, and grows; at the focus it goes to infinity. Throughout that motion the image stays real and inverted.
Object between the optical centre and $F$ of a convex lens: the image is virtual and erect, magnified, and on the same side as the object. The $v$ against $u$ graph for this range is a hyperbola-like branch on which $v$ is negative with $\lvert v\rvert > \lvert u\rvert$ , and $\lvert v\rvert$ rises without limit as $\lvert u\rvert$ approaches $f$ from below.

Power of a lens

Power measures the convergence or divergence a lens introduces in light falling on it. A ray parallel to the axis at unit distance from the optical centre is bent through an angle $\delta$ with $\tan\delta = h/f$ ; for $h=1$ and small $\delta$ , $\delta = 1/f$ . Hence

$P = \dfrac{1}{f}$

SI unit: the dioptre (D), $1\ \text{D} = 1\ \text{m}^{-1}$ . Definition to quote: one dioptre is the power of a lens of focal length one metre.
$P$ is positive for a converging lens and negative for a diverging lens. $P = +2.5$ D means a convex lens of $f = +40$ cm; $P = -4.0$ D means a concave lens of $f = -25$ cm. In centimetres, $P = 100/f$ .

Combination of thin lenses in contact

Diagram in words

Two thin lenses $A$ (focal length $f_1$ ) and $B$ (focal length $f_2$ ) in contact, their optical centres taken as coincident at $P$ . An object at $O$ beyond the focus of $A$ ; $A$ forms an image at $I_1$ , which acts as the object for $B$ ; $B$ forms the final image at $I$ .

Derivation

For lens $A$ : $\dfrac{1}{v_1} - \dfrac{1}{u} = \dfrac{1}{f_1}$
For lens $B$ , with $I_1$ as object: $\dfrac{1}{v} - \dfrac{1}{v_1} = \dfrac{1}{f_2}$
Adding, the $1/v_1$ terms cancel:

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$

If the pair is equivalent to a single lens of focal length $f$ , then $\dfrac1v - \dfrac1u = \dfrac1f$ , so

$\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2}, \qquad\text{and in general}\qquad \dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2} + \dfrac{1}{f_3} + \dots$

$P = P_1 + P_2 + P_3 + \dots \qquad m = m_1\,m_2\,m_3\dots$

The sum of powers is algebraic: convex lenses contribute positive powers, concave lenses negative. Combinations are used to obtain a desired focal length and magnification and to improve image sharpness, in cameras, microscopes and telescopes.
For lenses that are separated rather than in contact, do not use this formula: apply the lens formula to each lens in turn, treating each image as the object for the next lens, remembering that an image lying beyond the next lens is a virtual object with $u$ positive.

A lens in a liquid

$\dfrac{1}{f} = \left(\dfrac{n_{lens}}{n_{medium}} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

If $n_{lens} > n_{medium}$ : the lens keeps its character but the focal length increases (the power falls) — for example a convex lens of $f = 20$ cm in air ( $n_g = 1.5$ ) has $f = 50$ cm in a medium of $n = 1.25$ , still positive and still converging, just weaker.
If $n_{lens} < n_{medium}$ : the bracket is negative and the sign of $f$ flips — a biconvex lens of $n = 1.25$ in water of $n = 1.33$ diverges; a concave lens of $n = 1.5$ in a medium of $n = 1.65$ converges. A biconvex glass lens forming a real image in air forms a virtual image when immersed in a liquid of higher index.
If $n_{lens} = n_{medium}$ : $1/f = 0$ , $f\to\infty$ , the lens behaves as a plane sheet of glass and becomes invisible. A lens of $n = 1.47$ disappears in a liquid of $n = 1.47$ (glycerine, not water).
Wavelength: as $\lambda$ increases, $n$ decreases, so $(n-1)$ decreases, $f$ increases and $P$ decreases. Violet light is brought to a focus nearest the lens, red farthest.

Cutting a lens

Cut along the principal axis (a vertical cut through the axis, halving the aperture): neither radius changes, so $f$ and $P$ of each half are unchanged. Only the brightness of the image falls.
Cut perpendicular to the principal axis, separating the two curved faces: each half is a plano-lens with one radius replaced by $\infty$ . For an equiconvex lens of focal length $f$ , each half has focal length $2f$ and power $P/2$ ; for a biconcave lens of power $P$ , each plano-concave half has power $P/2$ . Putting the two halves back in contact restores the original, consistent with $\dfrac1f = \dfrac1F + \dfrac1F$ .

Numerical patterns CBSE repeats

Equiconvex or double convex lens with faces of equal radius: $R_1 = +R$ , $R_2 = -R$ , so $\dfrac1f = \dfrac{2(\mu-1)}{R}$ and $R = 2(\mu-1)f$ . With $\mu = 1.55$ and $f = 20$ cm, $R = 22$ cm. With $P = +5$ D and $\mu = 1.55$ , $R = 0.22$ m $= 22$ cm.
For $f = R$ in a biconvex lens: $\dfrac1R = (\mu-1)\dfrac2R$ gives $\mu = 1.5$ .
Equiconcave lens with $f = -\dfrac{3R}{4}$ : $R_1 = -R$ , $R_2 = +R$ , so $\dfrac{-4}{3R} = (\mu-1)\left(-\dfrac2R\right)$ and $\mu = 5/3$ .
Plano-convex lens, plane face first: $R_1 = \infty$ , $R_2 = -R$ , so $f = \dfrac{R}{\mu-1}$ . With $R = 20$ cm and $\mu = 1.5$ , $f = 40$ cm; an object at 30 cm then gives $v = -120$ cm, a virtual erect image magnified 4 times.
Concave lens of $\mu = 1.5$ , $f = -20$ cm in air, immersed in water of $\mu = 4/3$ : relative index $1.125$ , so $f_{water} = -80$ cm.
Virtual object. A beam converging towards a point $P$ is intercepted by a convex lens of $f = +10$ cm placed 15 cm before $P$ : $u = +15$ cm, so $\dfrac1v = \dfrac1{10} + \dfrac1{15} = \dfrac16$ and $v = +6$ cm.
Same-size real image: object and image both at $2f$ , so the object-to-screen distance is $4f$ and $f = d/4$ .
Displacement method. Screen 80 cm from the object, two lens positions 20 cm apart: $x = 30$ cm gives $\dfrac1f = \dfrac1{50} + \dfrac1{30}$ , so $f = 18.75$ cm. Equivalently $f = \dfrac{D^2-d^2}{4D} = \dfrac{6400-400}{320} = 18.75$ cm.
Lens plus plane mirror in contact, image coinciding with the object: the refracted rays must strike the mirror normally, so they are parallel, so the object is at the focus — $f$ equals the given object distance.
Lens plus convex mirror, final image coinciding with the object: the rays must fall normally on the mirror, so the image formed by the lens alone must be at the mirror's centre of curvature. With $u = -12$ cm, $f = 10$ cm the lens gives $v = 60$ cm; if the mirror is 10 cm from the lens, $R = 50$ cm and $f_{mirror} = 25$ cm.
Lens–mirror numericals in general: apply the lens formula, then shift the origin to the mirror (an image lying behind the mirror is a virtual object with $u$ positive), then apply the mirror formula.
Lens–lens separated. Object 36 cm before a convex lens of $f = 24$ cm gives $v_1 = 72$ cm; with a concave lens of $f = -18$ cm placed 45 cm away, $u_2 = +27$ cm gives $v_2 = -54$ cm, and $m = m_1m_2 = (-2)\times(-2) = 4$ .
Liquid-lens experiment. An equiconvex lens of radius $R$ rests on a liquid film on a plane mirror. The needle coincides with its image at distance $x$ with the liquid and at $y$ without it. Then $\dfrac1x = \dfrac1y + \dfrac1{f_2}$ , so $f_2 = \dfrac{xy}{y-x}$ ; the liquid film is a plano-concave lens with $R_1 = -R$ , $R_2 = \infty$ , so $\dfrac1{f_2} = -\dfrac{\mu_l-1}{R}$ and

$\mu_l = 1 + \dfrac{(x-y)R}{xy}$

Composite lens whose two halves have different refractive indices: each half has its own focal length, so two images are formed, not one.
A real image exists in space whether or not a screen is there; the screen only diffuses the light so the eye can see it.

Standard results

$\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ for one surface; apply it twice and add for a lens.
$\dfrac1f=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ ; $\dfrac1v-\dfrac1u=\dfrac1f$ ; $m=\dfrac vu$ ; $P=\dfrac1f$ in dioptre.
Lenses in contact: focal lengths combine as reciprocals, powers add, magnifications multiply.
The character of a lens depends on the lens material relative to its surroundings, not on its shape alone.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
A ray of light falls on a transparent sphere with centre $C$ as shown in the figure (the ray strikes the sphere's surface, refracts, crosses to the far surface and emerges parallel to the diameter line $AB$ , having entered making some angle of incidence at the first surface). The ray emerges from the sphere parallel to the line $AB$ . Find the angle of refraction at $A$ if the refractive index of the material of the sphere is $\sqrt3$ .
1 markCBSE Class 12 Physics — Very Short Answer · Foreign 2014, U · Section A, 1 markAnswer 1 ↓
2
A ray of light falls on a transparent sphere of $n=\sqrt3$ at an angle of incidence $60°$ with the diameter $AB$ of the sphere having centre $C$ . The ray emerges from the sphere parallel to the line $AB$ . Find the angle of emergence.
2 marksCBSE Class 12 Physics — SA · 2021 · Section B, 2 marksAnswer 2 ↓
3
Two transparent media of refractive indices $n_1$ and $n_2$ are separated by a spherical transparent surface. The rays of light incident on the surface get refracted into the medium on the other side. The laws of refraction are valid at each point of the spherical surface. A lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. The focal length of a lens is determined by the radii of curvature ( $R_1$ and $R_2$ ) of its two surfaces and the refractive index ( $n$ ) of the medium of the lens with respect to the surrounding medium. Depending on $R_1$ and $R_2$ a lens behaves as a diverging or a converging lens. The ability of a lens to diverge or converge a beam of light incident on it defines its power. (i) An object is placed at the point $B$ as shown in the figure (refraction at a single spherical surface separating media $n_1$ and $n_2$ , with the object on the $n_1$ side, at distance $u$ from the pole, and the surface's centre of curvature at distance $R$ ; the object distance $u$ and image distance $v$ are related as: (a) $\dfrac1v-\dfrac1u=\left(\dfrac{n_2-n_1}{n_1}\right)\dfrac1R$ (b) $\dfrac1v-\dfrac1u=\left(\dfrac{n_1-n_2}{n_2}\right)\dfrac1R$ (c) $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ (d) $\dfrac{n_1}{v}-\dfrac{n_2}{u}=\dfrac{n_1-n_2}{R}$ (ii) A point object is placed in air at a distance 'R' in front of a convex spherical refracting surface of radius of curvature R. If the medium on the other side of the surface is glass, then the image is: (a) real and formed in glass. (b) real and formed in air. (c) virtual and formed in glass. (d) virtual and formed in air. (iii) An object is kept at $2F$ in front of an equi-convex lens. The image formed is: (a) real and of the size of the object. (b) virtual and of the size of the object. (c) real and enlarged. (d) virtual and diminished. (iv) A thin converging lens of focal length 10 cm and a thin diverging lens of focal length 20 cm are placed coaxially in contact. The power of the combination is: (a) $-5$ D (b) $+5$ D (c) $+15$ D (d) $-15$ D (v) An equi-concave lens of focal length 'f' is cut into two identical parts along the dotted line as shown in the figure (a vertical cut through the middle, along the axis of symmetry, into two half-aperture pieces, each still concave-concave in cross-section but with radii $R$ marked on the two curved faces). The focal length of each part will be: (a) $\dfrac f4$ (b) $\dfrac f2$ (c) $f$ (d) $2f$
5 marksCBSE Class 12 Physics — Case · Term II 2021-22 · Section E, 5 marksAnswer 3 ↓
4
An object approaches a converging lens with a uniform speed of 5 m/s and stops at the focus. How will the image move with respect to the lens? Specify its nature.
1 markCBSE Class 12 Physics — Very Short Answer · 2021 · Section A, 1 markAnswer 4 ↓
5
A biconvex lens made up of glass of refractive index 1.5 forms a real image of an object in air. If the whole set-up were immersed in a liquid of refractive index 1.7, how will the nature of the image be affected?
1 markCBSE Class 12 Physics — Very Short Answer · 2020 · Section A, 1 markAnswer 5 ↓
6
A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens?
1 markCBSE Class 12 Physics — Very Short Answer · Delhi 2015 · Section A, 1 markAnswer 6 ↓
7
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.
1 markCBSE Class 12 Physics — Very Short Answer · AI 2014, Ap · Section A, 1 markAnswer 7 ↓
8
Using lens maker's formula, derive the thin lens formula $\dfrac1f=\dfrac1v-\dfrac1u$ .
2 marksCBSE Class 12 Physics — SA · 2020 · Section B, 2 marksAnswer 8 ↓
9
The focal length of an equi-concave lens is $3/4$ times the radius of curvature of its surfaces. Find the refractive index of the material of the lens. Under what condition will this lens behave as a converging lens?
2 marksCBSE Class 12 Physics — SA · 2020 · Section B, 2 marksAnswer 9 ↓
10
A screen is placed 80 cm from an object. The image of the object on the screen is formed by a convex lens placed between them at two different locations separated by a distance of 20 cm. Determine the focal length of the lens.
2 marksCBSE Class 12 Physics — SA · 2/5, 2020, Ev · Section B, 2 marksAnswer 10 ↓
11
A beam of light converges at a point $P$ . Now a convex lens is placed in the path of the convergent beam at 15 cm from point $P$ . At what point does the beam converge if the convex lens has a focal length 10 cm?
2 marksCBSE Class 12 Physics — SA · AI 2019 · Section B, 2 marksAnswer 11 ↓
12
A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens?
3 marksCBSE Class 12 Physics — SA · Delhi 2014, An · Section C, 3 marksAnswer 12 ↓
13
(a) An object is placed in front of a converging lens. Obtain the conditions under which the magnification produced by the lens is (i) negative and (ii) positive. (b) A point object is placed at $O$ , at a distance $3r$ in front of a glass sphere of radius $r$ , as shown in the figure (a sphere of radius $r$ centred at $C$ , with the object $O$ on the axis at distance $3r$ from the near surface's pole). Show the formation of image by the sphere.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marksAnswer 13 ↓
14
A double convex lens is made of a glass of refractive index 1.55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20 cm.
3 marksCBSE Class 12 Physics — SA · 1/3, AI 2017, Ap · Section C, 3 marksAnswer 14 ↓
15
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the image formation and determine the position and nature of the final image formed by the combined system.
3 marksCBSE Class 12 Physics — SA · AI 2014 · Section C, 3 marksAnswer 15 ↓
16
A convex lens of focal length 20 cm is placed coaxially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image.
4 marksCBSE Class 12 Physics — Long Answer · AI 2014 · Section E, 4 marksAnswer 16 ↓
17
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept 15 cm apart. A point object is placed 40 cm in front of the convex lens. Find the position of the image formed by this combination. Draw the ray diagram showing the image formation.
4 marksCBSE Class 12 Physics — Long Answer · AI 2014, An · Section E, 4 marksAnswer 17 ↓
18
A lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. Considering image formation by a single spherical surface successively at the two surfaces of a lens, the lens maker's formula is obtained. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. This formula helps us obtain a relation between $u$ , $v$ and $f$ for a lens. Lenses form images of objects and they are used in a number of optical devices, for example microscopes and telescopes. (i) An object $AB$ is kept in front of a composite convex lens made of two different media of refractive indices $n_1$ and $n_2$ (the top half of the lens has refractive index $n_1$ , the bottom half $n_2$ ), as shown in the figure. Will the lens produce one image? If not, explain. (ii) A real image of an object formed by a convex lens is observed on a screen. If the screen is removed, will the image still be formed? Explain. (iii) A double convex lens is made of glass of refractive index 1.55 with both faces of the same radius of curvature. Find the radius of curvature required if the focal length is 20 cm. OR (iii) Two convex lenses $A$ and $B$ of focal lengths 15 cm and 10 cm respectively are placed coaxially 'd' distance apart. A point object is kept at a distance of 30 cm in front of lens $A$ . Find the value of 'd' so that the rays emerging from lens $B$ are parallel to its principal axis.
5 marksCBSE Class 12 Physics — Long Answer · Section E, 5 marksAnswer 18 ↓
19
An object is placed 30 cm in front of a plano-convex lens with its spherical surface of radius of curvature 20 cm. If the refractive index of the material of the lens is 1.5, find the position and nature of the image formed.
2 marksCBSE Class 12 Physics — SA · 2/5, 2020, Ev · Section B, 2 marksAnswer 19 ↓
20
(a) Derive lens maker's formula for a biconvex lens. (b) A point object is placed at a distance of 12 cm on the principal axis of a convex lens of focal length 10 cm. A convex mirror is placed coaxially on the other side of the lens at a distance of 10 cm. If the final image coincides with the object, sketch the ray diagram and find the focal length of the convex mirror.
5 marksCBSE Class 12 Physics — Long Answer · 2020 · Section E, 5 marksAnswer 20 ↓
21
(a) Derive the mathematical relation between refractive indices $n_1$ and $n_2$ of two media and radius of curvature $R$ for refraction at a convex spherical surface. Consider the object to be a point source lying on the principal axis in rarer medium of refractive index $n_1$ and a real image formed in the denser medium of refractive index $n_2$ . Hence, derive lens maker's formula. (b) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of the light source from the glass surface is 100 cm. At what position is the image formed?
5 marksCBSE Class 12 Physics — Long Answer · AI 2016 · Section E, 5 marksAnswer 21 ↓
22
(a) A point object 'O' is kept in a medium of refractive index $n_1$ in front of a convex spherical surface of radius of curvature $R$ which separates the second medium of refractive index $n_2$ from the first one, as shown in the figure. Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of $n_1$ , $n_2$ and $R$ . (b) (i) When the image formed above acts as a virtual object for a concave spherical surface separating the medium $n_2$ from $n_1$ ( $n_2>n_1$ ), draw this ray diagram and write the similar (similar to (a)) relation. (ii) Hence, obtain the expression for the lens maker's formula.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2015, Ev · Section E, 5 marksAnswer 22 ↓
23
Draw a ray diagram showing the formation of the image by a point object on the principal axis of a spherical convex surface separating two media of refractive indices $n_1$ and $n_2$ , when a point source is kept in the rarer medium of refractive index $n_1$ . Derive the relation between object and image distance in terms of refractive index of the medium and radius of curvature of the surface. Hence obtain the expression for lens-maker's formula in the case of a thin convex lens.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2014C · Section E, 5 marksAnswer 23 ↓
24
(a) A point object is placed in front of a double convex lens (of refractive index $n=n_2/n_1$ with respect to air) with its spherical faces of radii of curvature $R_1$ and $R_2$ . Show the path of rays due to refraction at first and subsequently at the second surface to obtain the formation of the real image of the object. Hence obtain the lens-maker's formula for a thin lens. (b) A double convex lens having both faces of the same radius of curvature has refractive index 1.55. Find out the radius of curvature of the lens required to get the focal length of 20 cm.
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 24 ↓
25
What is the power of an equiconvex lens of refractive index $n_2$ dipped in a liquid of refractive index $n_1$ , where $n_1<n_2$ ?
1 markCBSE Class 12 Physics — Very Short Answer · 2021 · Section A, 1 markAnswer 25 ↓
26
(i) Define SI unit of power of a lens. (ii) A plano convex lens is made of glass of refractive index 1.5. The radius of curvature of the convex surface is 25 cm. (a) Calculate the focal length of lens. (b) If an object is placed 50 cm in front of the lens, find the nature and position of the image formed.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marksAnswer 26 ↓
27
The lens maker's formula is useful to design lenses of desired focal lengths using surfaces of suitable radii of curvature. The focal length also depends on the refractive index of the material of the lens and the surrounding medium. The refractive index depends on the wavelength of the light used. The power of a lens is related to its focal length. Answer the following questions based on the above: (a) How will the power of a lens be affected with an increase of wavelength of light? (b) The radius of curvature of two surfaces of a convex lens is $R$ each. For what value of $\mu$ of its material will its focal length become equal to $R$ ? (c) The focal length of a concave lens of $\mu=1.5$ is 20 cm in air. It is completely immersed in water of $\mu=\dfrac43$ . Calculate its focal length in water. OR (c) An object is placed in front of a lens which forms its erect image of magnification 3. The power of the lens is 5D. Calculate the distance of the object and the image from the lens.
4 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 4 marksAnswer 27 ↓
28
A lens is a portion of a transparent medium bounded by two surfaces and one of these surfaces is essentially spherical. It is used to converge or diverge the light incident on it. Power of a lens is the measure of its ability to converge or diverge the light incident on it. Power of a lens depends on the refractive index of the material of the lens with respect to the surrounding medium and the radii of curvature of its two surfaces. (a) The power of a convex lens (i) increases when the lens is dipped in water. (ii) increases when the wavelength of incident light decreases. (iii) decreases when another convex lens is placed in contact with it. (iv) increases when the lens is cut into two identical plano-convex lenses. (b) The focal length of a concave lens is 40 cm. The power of the lens is (i) $0.025$ D (ii) $2.5$ D (iii) $-0.025$ D (iv) $-2.5$ D (c) The focal length of a concave lens ( $\mu_g=1.5$ ) in air is 20 cm. What should be the refractive index of the surrounding medium so that the lens behaves as a converging lens of focal length 60 cm? (i) 1.4 (ii) 1.8 (iii) 1.7 (iv) 1.2 (d) Beams of red light, blue light, yellow light and violet light are incident on a convex lens, one-by-one. Which one of them converges nearest to the lens? (i) Blue light (ii) Violet light (iii) Red light (iv) Yellow light (e) A beam of light coming parallel to the principal axis of a convex lens $L_1$ of focal length 15 cm is incident on it. Another convex lens $L_2$ of focal length 25 cm is placed coaxially at a distance $d$ from $L_1$ . For the final image to be formed at infinity, the value of $d$ should be: (i) 10 cm (ii) 15 cm (iii) 25 cm (iv) 40 cm
5 marksCBSE Class 12 Physics — Long Answer · 2022 C · Section E, 5 marksAnswer 28 ↓
29
An equi-convex lens of focal length 'f' is cut into two identical plane convex lenses. How will the power of each part be related to the focal length of the original lens? A double convex lens of $+5$ D is made of glass of refractive index 1.55 with both faces of equal radii of curvature. Find the value of its radius of curvature.
2 marksCBSE Class 12 Physics — SA · Foreign 2015, An · Section B, 2 marksAnswer 29 ↓
30
A symmetric biconvex lens of radius of curvature $R$ and made of glass of refractive index 1.5, is placed on a layer of liquid placed on top of a plane mirror. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be $x$ . On removing the liquid layer and repeating the experiment, the distance is found to be $y$ . Obtain the expression for the refractive index of the liquid in terms of $x$ and $y$ .
3 marksCBSE Class 12 Physics — SA · 2018, Cr · Section C, 3 marksAnswer 30 ↓
31
In the following diagram, an object 'O' is placed 15 cm in front of a convex lens $L_1$ of focal length 20 cm; the two lenses $L_1$ and $L_2$ are 20 cm apart, and the final image is formed at $I$ at a distance of 80 cm from the second lens $L_2$ . Find the focal length of the lens $L_2$ .
4 marksCBSE Class 12 Physics — Long Answer · Foreign 2016, An · Section E, 4 marksAnswer 31 ↓
32
(a) Using the ray diagram for a system of two lenses of focal lengths $f_1$ and $f_2$ in contact with each other, show that the two lens system can be regarded as equivalent to a single lens of focal length $f$ , where $\dfrac1f=\dfrac1{f_1}+\dfrac1{f_2}$ . Also write the relation for the equivalent power of the lens combination. (b) Determine the position of the image formed by the lens combination given in the figure: three thin lenses of focal lengths $f_1=+10$ cm, $f_2=-10$ cm and $f_3=+30$ cm placed coaxially, with an object 30 cm in front of the first lens, a gap of 5 cm between the first and second lens, and a gap of 10 cm between the second and third lens.
5 marksCBSE Class 12 Physics — Long Answer · AI 2019, Ev · Section E, 5 marksAnswer 32 ↓
33
Draw a ray diagram to show the image formation by a combination of two thin convex lenses in contact. Obtain the expression for the power of this combination in terms of the focal lengths of the lenses.
5 marksCBSE Class 12 Physics — Long Answer · 2/5, AI 2015 · Section E, 5 marksAnswer 33 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

34
A window is provided in the middle of a wall. Its image is obtained on the opposite wall at a distance 'd' from it using a lens. If the window and its image are of the same size, then the focal length of the lens used is
- (a)
  $+\dfrac d4$
- (b)
  $+\dfrac d2$
- (c)
  $-\dfrac d4$
- (d)
  $-\dfrac d2$
1 markCBSE Class 12 Physics — Multiple Choice · 2020 · Section A, 1 markAnswer 34 ↓
35
A biconvex lens of glass having refractive index 1.47 is immersed in a liquid. It becomes invisible and behaves as a plane glass plate. The refractive index of the liquid is
- (a)
  1.47
- (b)
  1.62
- (c)
  1.33
- (d)
  1.51
1 markCBSE Class 12 Physics — Multiple Choice · 2020, U · Section A, 1 markAnswer 35 ↓
36
A biconcave lens of power $P$ vertically splits into two identical plano-concave parts. The power of each part will be
- (a)
  $2P$
- (b)
  $P/2$
- (c)
  $P$
- (d)
  $P/\sqrt2$
1 markCBSE Class 12 Physics — Multiple Choice · 2020, U · Section A, 1 markAnswer 36 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
Let the ray meet the sphere at $A$ with angle of incidence $i$ and angle of refraction $r$ . The inward normal at $A$ lies along the diameter $AB$ , so the refracted ray leaves $A$ at angle $r$ to $AB$ . The refracted ray is the chord $AD$ to the far surface. Since $CA=CD=$ radius, triangle $ACD$ is isosceles, so the chord meets the normal $CD$ at $D$ at the same angle $r$ ; the ray therefore emerges at $D$ at angle $i$ to $CD$ . The outward normal $CD$ is inclined to $AB$ at the exterior angle $2r$ , and the emergent ray lies at $i$ on the far side of it, so the emergent ray is inclined to $AB$ at $(2r-i)$ . The emergent ray is given as parallel to $AB$ , hence $\;2r-i=0\;\Rightarrow\;i=2r$ Substituting into Snell's law at $A$ : $\;\sin i=n\sin r\;\Rightarrow\;\sin 2r=\sqrt3\,\sin r\;\Rightarrow\;2\sin r\cos r=\sqrt3\,\sin r$ $\;\Rightarrow\;\cos r=\dfrac{\sqrt3}{2}\;\Rightarrow\;r=30°$ (and $i=60°$ ), so the angle of refraction at $A$ is $30°$ .
[Note on this record: the earlier version of this solution took $i=60°$ from the figure as a standard result. It is not an assumption — $i=2r$ follows from the sphere's own geometry, and the derivation above needs no figure. The answer is unchanged.]
2.↑ back to the question
At the first surface, by Snell's law: $\;n=\dfrac{\sin i}{\sin r}\;\Rightarrow\;\sqrt3=\dfrac{\sin60°}{\sin r}\;\Rightarrow\;\sin r=\dfrac{\sin60°}{\sqrt3}=\dfrac{\sqrt3/2}{\sqrt3}=\dfrac12\;\Rightarrow\;r=30°$ By the symmetry of the sphere (isosceles triangle formed by the two radii to the points of incidence and emergence, and the chord), the angle of incidence at the second (exit) surface, measured from the local normal (which is along the radius), equals $r=30°$ as well. Applying Snell's law at the exit surface (going from the denser medium to air): $\;\dfrac{1}{n}=\dfrac{\sin(30°)}{\sin e}\;\Rightarrow\;\sin e=n\sin30°=\sqrt3\times\dfrac12=\dfrac{\sqrt3}{2}$ $\Rightarrow e=60°$
3.↑ back to the question
(i) (d): Using the standard derivation for refraction at a spherical surface with $O$ at distance $u$ in medium $n_1$ and image at distance $v$ in medium $n_2$ , the relevant relation for the labelling shown in this figure works out to $\dfrac{n_1}{v}-\dfrac{n_2}{u}=\dfrac{n_1-n_2}{R}$ , consistent with option (d) for the object position and sign convention shown. (ii) (d): With the point object in air (rarer, $n_1=1$ ) and glass ( $n_2$ ) on the other side of a convex surface, using $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ with $u=-R$ : $\;\dfrac{\mu_g}{v}-\dfrac{1}{-R}=\dfrac{\mu_g-1}{R}\;\Rightarrow\;\dfrac{\mu_g}{v}=\dfrac{\mu_g-1}{R}-\dfrac1R=\dfrac{\mu_g-2}{R}$ With $\mu_g=1.5$ : $\;\dfrac{1.5}{v}=\dfrac{1.5-2}{R}=-\dfrac{0.5}{R}\;\Rightarrow\;v=-3R$ , i.e. $v$ is negative, so the image forms on the same side as the object — it is virtual, and formed in the air. Hence option (d). (iii) (a): When an object is kept at $2F$ of an equi-convex (thin) lens, the image forms at $2F$ on the other side, real and of the same size as the object. Hence option (a). (iv) (b): $\dfrac1f=\dfrac1{f_1}+\dfrac1{f_2}=\dfrac1{10}-\dfrac1{20}=\dfrac1{20}\;\Rightarrow\;f=+20$ cm, so $P=\dfrac{100}{f(\text{cm})}=\dfrac{100}{20}=+5$ D. Hence option (b). (v) (c): Cutting the lens vertically along the optical axis (as shown) only reduces its aperture — it does not change either radius of curvature, so by $\dfrac1f=(\mu-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ , the focal length of each half remains the same as the original, $f$ . Hence option (c).
4.↑ back to the question
As the object moves from far away towards the focus of the converging lens, the image distance $v$ increases (moves farther from the lens), and moreover it moves away from the lens with increasing speed as the object approaches the focus (since $v\to\infty$ as $u\to f$ ). Throughout this motion (object outside the focus), the image is real, inverted, and becomes progressively larger (magnified) as the object approaches the focus.
5.↑ back to the question
When the convex lens (of refractive index 1.5) is immersed in a liquid of higher refractive index (1.7), the lens behaves like a diverging (concave) lens, since the surrounding medium is now optically denser than the lens material. As a result, the image, which was real when the lens acted as a converging lens in air, will now become virtual.
6.↑ back to the question
The focal length of a concave lens in air is negative. Using the lens-maker's formula, $\dfrac1f=\left(\dfrac{\mu_l}{\mu_m}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ , with $\mu_l=1.5$ (lens) and $\mu_m=1.65$ (medium). Since $\mu_l<\mu_m$ , the factor $\left(\dfrac{\mu_l}{\mu_m}-1\right)$ is negative, which flips the sign of the focal length compared to when the lens is in air — the focal length of the given (concave) lens becomes positive. Hence, when immersed in this denser medium, the concave lens behaves as a convex (converging) lens.
7.↑ back to the question
The lens will act as a diverging lens, because the refractive index of the surrounding medium (water, 1.33) is greater than that of the lens material (1.25) — the factor $\left(\dfrac{\mu_{lens}}{\mu_{water}}-1\right)$ in the lens-maker's formula becomes negative, reversing the usual converging behaviour of a biconvex shape.
8.↑ back to the question
Consider a thin lens with surfaces of radii $R_1$ and $R_2$ , refractive index $n_2$ , placed in a medium of refractive index $n_1$ . For refraction at the first surface $XP_1Y$ alone (ignoring the second surface for now), with the object at $O$ (distance $u$ ) and a virtual intermediate image at $I'$ (distance $v'$ ), light travels from $n_1$ to $n_2$ : $\dfrac{n_2}{v'}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R_1}\qquad\text{...(i)}$ For refraction at the second surface $XP_2Y$ , the point $I'$ acts as a virtual object, and light now travels from $n_2$ to $n_1$ , forming the final image at $I$ (distance $v$ ): $\dfrac{n_1}{v}-\dfrac{n_2}{v'}=\dfrac{n_1-n_2}{R_2}\qquad\text{...(ii)}$ Adding equations (i) and (ii), the $n_2/v'$ terms cancel: $\;\dfrac{n_1}{v}-\dfrac{n_1}{u}=(n_2-n_1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ Dividing throughout by $n_1$ : $\;\dfrac1v-\dfrac1u=\left(\dfrac{n_2}{n_1}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\qquad\text{...(iii)}$ When the object is at infinity ( $u\to-\infty$ ), the image forms at the focus, $v=f$ : $\;\dfrac1f=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\qquad\text{...(iv)}$ — this is the lens-maker's formula. Comparing equations (iii) and (iv), the right-hand sides are identical, so: $\;\dfrac1v-\dfrac1u=\dfrac1f$ — this is the thin lens formula, valid for a biconvex (and any thin) lens.
9.↑ back to the question
For an equi-concave lens, $R_1=-R$ and $R_2=+R$ (both surfaces concave, magnitude $R$ ), and $f=-\dfrac{3R}{4}$ (concave lens, focal length negative). From the lens-maker's formula: $\;\dfrac1f=(\mu-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\;\Rightarrow\;\dfrac{-4}{3R}=(\mu-1)\left(\dfrac1{-R}-\dfrac1R\right)=(\mu-1)\left(\dfrac{-2}{R}\right)$ $\Rightarrow\dfrac{4}{3R}=(\mu-1)\dfrac2R\;\Rightarrow\;\mu-1=\dfrac23\;\Rightarrow\;\mu=\dfrac53$ This lens will behave as a converging lens if $(\mu-1)<0$ , i.e. if $\mu<1$ — which happens if the lens is placed in a surrounding medium of refractive index greater than that of the lens material (so that the relative refractive index of the lens material with respect to the medium is less than 1).
10.↑ back to the question
This is the displacement method. Let the lens, in its first position, be at a distance $x$ from the object; then $u=-x$ , $v=80-x$ , and using the lens formula with the (unknown, same) focal length $f$ : $\dfrac1f=\dfrac1{80-x}+\dfrac1x\qquad\text{...(i)}$ In the second position, the lens has moved by 20 cm, so $u=-(x+20)$ , $v=60-x$ : $\dfrac1f=\dfrac1{60-x}+\dfrac1{x+20}\qquad\text{...(ii)}$ Equating (i) and (ii): $\;\dfrac1{80-x}+\dfrac1x=\dfrac1{60-x}+\dfrac1{x+20}$ Solving this equation gives $x=30$ cm. Substituting back into (i): $\;\dfrac1f=\dfrac1{80-30}+\dfrac1{30}=\dfrac1{50}+\dfrac1{30}=\dfrac{3+5}{150}=\dfrac{8}{150}$ $\Rightarrow f=\dfrac{150}{8}=18.75$ cm.
11.↑ back to the question
Since the beam is converging towards $P$ when the lens is inserted, $P$ acts as a virtual object for the lens, at $u=+15$ cm (on the far side, in the direction the light is travelling), and $f=+10$ cm (convex lens). Using the lens formula: $\;\dfrac1v-\dfrac1u=\dfrac1f\;\Rightarrow\;\dfrac1v=\dfrac1f+\dfrac1u=\dfrac1{10}+\dfrac1{15}=\dfrac{3+2}{30}=\dfrac{5}{30}=\dfrac16$ $\Rightarrow v=6$ cm. So the beam now converges at a point 6 cm from the lens (on the far side), instead of at $P$ .
12.↑ back to the question
Since the final image coincides with the object itself, the rays refracted by the lens must fall normally on the plane mirror (so that they retrace their exact path back through the lens to the object). For rays to strike a plane mirror normally (i.e. travel parallel to the principal axis just before hitting the flat mirror), the lens must have converted the diverging rays from the object into a parallel beam — this happens precisely when the object is placed at the focus of the lens. Hence the focal length of the lens equals the given object distance: $f=20$ cm.
13.↑ back to the question
(a) Converging lens means a convex lens. Consider an object $AB$ beyond the focus, forming a real, inverted image $A'B'$ . By similar triangles $\triangle ABC\sim\triangle A'B'C$ (where $C$ is the optical centre): $\;\dfrac{A'B'}{AB}=\dfrac{CB'}{CB}$ (i) When the object lies beyond the focus $F$ (i.e. $u<-f$ ), the image formed is real and inverted, giving magnification $m=\dfrac{h_2}{h_1}=\dfrac{-v}{-u}=-\dfrac vu$ , and since both $u,v$ have the same sign convention here with $v$ positive (real image) and $u$ negative, $m$ works out negative — magnification is negative whenever the image is real (object beyond $F$ ). (ii) When the object lies between the focus $F$ and the optical centre (i.e. $0<|u|<f$ ), the image formed is virtual, erect and enlarged, on the same side as the object; here $v$ is also negative, so $m=-v/u$ works out positive — magnification is positive whenever the image is virtual (object between the lens and its focus). (b) The image is formed by two successive refractions, one at each curved surface of the sphere. At the first (near) surface, the object $O$ (in air, at distance $3r$ from the pole, with the surface's centre of curvature at the sphere's centre $C$ , radius $r$ ) forms an intermediate (real) image inside the glass sphere, closer to the far surface, found using $\dfrac{n_2}{v_1}-\dfrac{n_1}{u_1}=\dfrac{n_2-n_1}{R_1}$ with $u_1=-3r$ , $R_1=+r$ . This intermediate image then acts as the object for refraction at the second (far) surface of the sphere, where light passes from glass back into air; applying $\dfrac{n_1}{v}-\dfrac{n_2}{u_2}=\dfrac{n_1-n_2}{R_2}$ (with $R_2=-r$ and $u_2$ measured from the far pole) gives the position of the final image $I$ , which forms outside the sphere, further along the axis at a distance $r$ from the far surface's pole for this specific configuration (a standard result for an object at $3r$ from a sphere of radius $r$ ). Thus, the sphere forms a real image $I$ of the point object $O$ , located outside the sphere on the far side, found by combining the two refractions in sequence.
14.↑ back to the question
For a double convex lens with both faces of equal radius $R$ : $R_1=+R$ , $R_2=-R$ ; $\mu=1.55$ , $f=20$ cm. Using the lens-maker's formula: $\;\dfrac1f=(\mu-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\;\Rightarrow\;\dfrac1{20}=(1.55-1)\left(\dfrac1R+\dfrac1R\right)=0.55\times\dfrac2R$ $\Rightarrow R=0.55\times2\times20=22\text{ cm}$
15.↑ back to the question
For the convex lens: $u=-60$ cm, $f=+20$ cm. Using $\dfrac1v-\dfrac1u=\dfrac1f$ : $\;\dfrac1v=\dfrac1f+\dfrac1u=\dfrac1{20}-\dfrac1{60}=\dfrac{3-1}{60}=\dfrac{2}{60}=\dfrac1{30}\;\Rightarrow\;v=+30$ cm. This image (30 cm from the lens) acts as a virtual object for the convex mirror, which is 15 cm from the lens: so the object distance for the mirror is $u=+(30-15)=+15$ cm (virtual object, behind the mirror), and $f_{mirror}=+\dfrac{20}{2}=+10$ cm. Using the mirror formula $\dfrac1v+\dfrac1u=\dfrac1f$ : $\;\dfrac1v=\dfrac1{10}-\dfrac1{15}=\dfrac{3-2}{30}=\dfrac1{30}\;\Rightarrow\;v=+30$ cm. The final image is formed at a distance of 30 cm from the convex mirror (equivalently 45 cm from the convex lens), on the same side as the incoming light — the final image formed is virtual.
16.↑ back to the question
For the convex lens: since the incident beam is parallel to the axis, $u=-\infty$ , $f=+20$ cm, so the image $I_1$ forms at the focus: $v=+20$ cm. This image $I_1$ (20 cm from the lens) lies in front of the concave mirror (which is 50 cm from the lens), so it acts as a real object for the mirror at a distance of $50-20=30$ cm; for the concave mirror, $u=-30$ cm, $f=-10$ cm. Using the mirror formula: $\;\dfrac1v+\dfrac1u=\dfrac1f\;\Rightarrow\;\dfrac1v=\dfrac1{-10}-\dfrac1{-30}=-\dfrac1{10}+\dfrac1{30}=\dfrac{-3+1}{30}=-\dfrac{2}{30}=-\dfrac1{15}$ $\Rightarrow v=-15$ cm. Hence the final image $I_2$ is formed by the concave mirror at a distance of 15 cm in front of it (i.e. between the mirror and the lens).
17.↑ back to the question
For the convex lens: $u=-40$ cm, $f=+20$ cm. $\;\dfrac1v=\dfrac1f+\dfrac1u=\dfrac1{20}-\dfrac1{40}=\dfrac{2-1}{40}=\dfrac1{40}\;\Rightarrow\;v=+40$ cm. This image forms 40 cm from the lens, i.e. $40-15=25$ cm beyond the convex mirror (which is 15 cm from the lens) — so it acts as a virtual object for the mirror at $u=+25$ cm; $f_{mirror}=+\dfrac{20}{2}=+10$ cm. Using the mirror formula: $\;\dfrac1v=\dfrac1f-\dfrac1u=\dfrac1{10}-\dfrac1{25}=\dfrac{5-2}{50}=\dfrac{3}{50}\;\Rightarrow\;v=\dfrac{50}{3}\approx16.67$ cm. So the final image is formed at $\dfrac{50}{3}\approx16.67$ cm ( $=\dfrac{50}{3}$ cm) to the right of (behind) the convex mirror, equivalently at $\dfrac{50}{3}+15=\dfrac{95}{3}\approx31.67$ cm to the right of the convex lens.
18.↑ back to the question
(i) No. Since the two halves of the composite lens have different refractive indices $n_1$ and $n_2$ , the lens effectively has two different focal lengths (one for each half, computed via the lens-maker's formula with $n_1$ and $n_2$ respectively). As a result, each half forms its own image at a different position, so two separate images are formed instead of one. (ii) Yes, the real image will still be formed at exactly the same point in space, even without a screen there. A real image represents an actual convergence of light rays at that point; the screen only makes this convergence visible to an observer — its presence or absence does not affect whether or where the rays converge. (iii) $R_1=+R$ , $R_2=-R$ , $\mu=1.55$ , $f=20$ cm. $\;\dfrac1{20}=(1.55-1)\left(\dfrac1R+\dfrac1R\right)=0.55\times\dfrac2R\;\Rightarrow\;R=0.55\times2\times20=22$ cm. OR (iii) For lens $A$ : $u_A=-30$ cm, $f_A=15$ cm. $\;\dfrac1{v_A}=\dfrac1{15}-\dfrac1{30}=\dfrac{2-1}{30}=\dfrac1{30}\;\Rightarrow\;v_A=30$ cm. For the rays to emerge from lens $B$ parallel to the principal axis, the intermediate image (from lens $A$ ) must lie exactly at the focus of lens $B$ — i.e. the object distance for lens $B$ must equal $f_B=10$ cm, on the correct (virtual-object) side. This intermediate image is at $30$ cm from lens $A$ ; for it to be at 10 cm in front of lens $B$ 's focus in the appropriate sense (i.e. lens $B$ placed such that this image is a virtual object at $u_B=-(d-30)$ , requiring the image to be at $d-30=f_B=10$ so that $v_B=\infty$ ), solving gives $d=40$ cm.
19.↑ back to the question
For the plano-convex lens: $R_1=\infty$ (plane face), $R_2=-20$ cm (curved face), $\mu=1.5$ . Using the lens-maker's formula: $\;\dfrac1f=(\mu-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=(1.5-1)\left(\dfrac1\infty-\dfrac1{-20}\right)=0.5\times\dfrac1{20}=\dfrac1{40}$ $\Rightarrow f=40$ cm. Given $u=-30$ cm. Using the lens formula: $\;\dfrac1v-\dfrac1u=\dfrac1f\;\Rightarrow\;\dfrac1v=\dfrac1{40}+\dfrac1{-30}=\dfrac1{40}-\dfrac1{30}=\dfrac{3-4}{120}=-\dfrac1{120}$ $\Rightarrow v=-120$ cm. The image is formed 120 cm from the lens, on the same side as the object; since $v$ is negative, the image is virtual and erect. Magnification: $\;m=\dfrac vu=\dfrac{-120}{-30}=4$ So the image is virtual, erect, and magnified 4 times the size of the object.
20.↑ back to the question
(a) [See the general derivation: for refraction at each of the lens's two surfaces in turn, combining $\dfrac{n_2-n_1}{R_1}=\dfrac{n_2}{v'}-\dfrac{n_1}{u}$ and $\dfrac{n_1-n_2}{R_2}=\dfrac{n_1}{v}-\dfrac{n_2}{v'}$ and adding, one obtains $\dfrac1v-\dfrac1u=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ , and setting $u\to-\infty$ , $v\to f$ , gives the lens-maker's formula $\dfrac1f=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ — see Q24/Q39 for the full step-by-step version of this derivation.] (b) Since the final image coincides with the object itself, the rays from the object, after refraction through the lens and reflection from the mirror, must retrace their original path. This requires the rays refracted by the lens to fall normally on the mirror — which happens only if these refracted rays are directed exactly towards the mirror's centre of curvature. So the image formed by the lens alone must coincide with the centre of curvature of the mirror. For the lens: $u=-12$ cm, $f=10$ cm. Using the lens formula: $\;\dfrac1v-\dfrac1{-12}=\dfrac1{10}\;\Rightarrow\;\dfrac1v=\dfrac1{10}-\dfrac1{12}=\dfrac{6-5}{60}=\dfrac1{60}$ $\Rightarrow v=60$ cm. So the lens forms an image 60 cm from the lens. Since the mirror is 10 cm from the lens, this image lies at a distance $R=v-d=60-10=50$ cm from the mirror — this must equal the mirror's radius of curvature for the rays to strike it normally. Hence $R=50$ cm, and the focal length of the convex mirror is $f_{mirror}=\dfrac R2=25$ cm.
21.↑ back to the question
(a) Let a point object $O$ lie in the rarer medium ( $n_1$ ) on the axis, with a ray $OA$ striking the convex spherical surface (centre $C$ , pole $P$ ) at $A$ , refracting into the denser medium ( $n_2$ ) and meeting the axis at the real image point $I$ . Let $\alpha,\beta,\gamma$ be the (small) angles that $OA$ , $IA$ and $CA$ make with the axis. In $\triangle OAC$ , the angle of incidence $i=\alpha+\gamma$ ; in $\triangle AIC$ , $\gamma=r+\beta$ , so the angle of refraction $r=\gamma-\beta$ . By Snell's law: $\;^1n_2=\dfrac{\sin i}{\sin r}\approx\dfrac ir=\dfrac{\alpha+\gamma}{\gamma-\beta}\;\Rightarrow\;n_2(\gamma-\beta)=n_1(\alpha+\gamma)\;\Rightarrow\;(n_2-n_1)\gamma=n_1\alpha+n_2\beta\qquad\text{...(i)}$ For paraxial rays ( $\alpha,\beta,\gamma$ small, and $P,N$ nearly coincident): $\;\alpha\approx\dfrac{AN}{PO},\quad\beta\approx\dfrac{AN}{PI},\quad\gamma\approx\dfrac{AN}{PC}$ Substituting into (i): $\;(n_2-n_1)\dfrac{AN}{PC}=n_1\dfrac{AN}{PO}+n_2\dfrac{AN}{PI}\;\Rightarrow\;\dfrac{n_2-n_1}{PC}=\dfrac{n_1}{PO}+\dfrac{n_2}{PI}$ With sign convention $PC=+R$ , $PO=-u$ , $PI=+v$ : $\;\dfrac{n_2-n_1}{R}=\dfrac{n_1}{-u}+\dfrac{n_2}{v}\;$ , i.e. $\;\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ This is the required relation for refraction at a convex spherical surface, object in the rarer medium, real image in the denser medium. For a thin lens, applying this at both surfaces in turn and combining (as in Q24) gives the lens-maker's formula $\dfrac1f=\left(\dfrac{n_2}{n_1}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ . (b) $R=20$ cm, $n_2=1.5$ , $n_1=1$ , $u=-100$ cm. $\;\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}\;\Rightarrow\;\dfrac{1.5}{v}+\dfrac1{100}=\dfrac{0.5}{20}$ $\dfrac{1.5}{v}=\dfrac1{40}-\dfrac1{100}=\dfrac{5-2}{200}=\dfrac{3}{200}\;\Rightarrow\;v=\dfrac{200}{3}\times1.5=100\text{ cm}$ So a real image is formed 100 cm away from the surface, on the other (glass) side.
22.↑ back to the question
(a) [Same derivation as Q37(a) above]: for a point object $O$ in medium $n_1$ , refracting at a convex surface of radius $R$ into medium $n_2$ ( $n_1<n_2$ ), the relation obtained is $\;\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}\qquad\text{...(i)}$ , with $u,v$ measured from the pole using the Cartesian sign convention. (b) (i) Now consider the image from (a) acting as a virtual object for a second, concave spherical surface (radius $R'$ ) which separates the denser medium $n_2$ (in which the virtual object now sits) from $n_1$ on the far side ( $n_2>n_1$ ). By the same method (now with light travelling from $n_2$ into $n_1$ across this second surface): $\;\dfrac{n_1}{v'}-\dfrac{n_2}{v}=\dfrac{n_1-n_2}{R'}\qquad\text{...(ii)}$ (ii) Adding equations (i) and (ii) (with appropriate relabelling for a thin lens, where the two surfaces are close together so the intermediate image distance $v$ is common to both), the $n_2$ terms cancel, giving $\;\dfrac{n_1}{v'}-\dfrac{n_1}{u}=(n_2-n_1)\left(\dfrac1R-\dfrac1{R'}\right)$ Dividing by $n_1$ and writing $R=R_1$ , $R'=R_2$ , $v'=v$ (final image): $\;\dfrac1v-\dfrac1u=\left(\dfrac{n_2}{n_1}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ Setting $u\to-\infty$ gives $v\to f$ , so $\;\dfrac1f=\left(\dfrac{n_2}{n_1}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ — the lens-maker's formula — and hence, generally, $\dfrac1v-\dfrac1u=\dfrac1f$ (the thin lens formula).
23.↑ back to the question
[Same derivation as Q37(a)]: for a point object $O$ in the rarer medium $n_1$ , on the axis, refracting at a convex spherical surface of radius $R$ into the denser medium $n_2$ , using the small-angle (paraxial) approximation for the angles subtended at the surface by the incident ray, the normal, and the refracted ray, and applying Snell's law, one obtains: $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ Applying this relation successively at the two surfaces of a thin convex lens (as in Q24/Q38) and adding, the intermediate-image term cancels, giving the thin lens formula $\dfrac1v-\dfrac1u=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ , and setting the object at infinity gives the lens-maker's formula $\dfrac1f=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ .
24.↑ back to the question
(a) [Same construction as Q24]: for refraction at the first surface (radius $R_1$ ) of the lens, with the object $O$ in air (refractive index $n_1$ ) forming an intermediate (virtual) image $I'$ inside the lens material ( $n_2$ ): $\;\dfrac{n_2-n_1}{R_1}=\dfrac{n_2}{v'}-\dfrac{n_1}{u}\qquad\text{...(i)}$ For refraction at the second surface (radius $R_2$ ), $I'$ acts as a virtual object, and the final real image $I$ forms back in air: $\;\dfrac{n_1-n_2}{R_2}=\dfrac{n_1}{v}-\dfrac{n_2}{v'}\qquad\text{...(ii)}$ Adding (i) and (ii), the $n_2/v'$ terms cancel: $\;\dfrac{n_1}{v}-\dfrac{n_1}{u}=(n_2-n_1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ Dividing by $n_1$ : $\;\dfrac1v-\dfrac1u=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ ; setting $u\to-\infty$ , $v\to f$ gives the lens-maker's formula, $\;\dfrac1f=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ . (b) $R_1=R$ , $R_2=-R$ , $\mu=1.55$ , $f=20$ cm: $\;\dfrac1{20}=(1.55-1)\left(\dfrac1R+\dfrac1R\right)=0.55\times\dfrac2R\;\Rightarrow\;R=0.55\times2\times20=22$ cm.
25.↑ back to the question
$P=\dfrac1f=\left(\dfrac{\mu_L}{\mu_S}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ , where $\mu_L=n_2$ (lens) and $\mu_S=n_1$ (surrounding liquid), and for an equiconvex lens $R_1=+R$ , $R_2=-R$ . $P=\left(\dfrac{n_2}{n_1}-1\right)\left(\dfrac1R+\dfrac1R\right)=\left(\dfrac{n_2-n_1}{n_1}\right)\dfrac2R$
26.↑ back to the question
(i) The SI unit of power of a lens is the dioptre (D). A power of one dioptre is the power of a lens whose focal length is 1 metre, i.e. $1\text{ D}=1\text{ m}^{-1}$ . (ii) Given: radius of curvature of the convex surface $=25$ cm, $\mu=1.5$ . (a) For this plano-convex lens (as oriented and labelled in the figure), $\;f=\dfrac{R}{\mu-1}=\dfrac{-25}{1.5-1}=\dfrac{-25}{0.5}=-50\text{ cm}$ (b) $u=-50$ cm. Using the lens formula: $\;\dfrac1v=\dfrac1f+\dfrac1u=\dfrac1{-50}+\dfrac1{-50}=-\dfrac2{50}=-\dfrac1{25}$ $\Rightarrow v=-25$ cm. Magnification $m=\dfrac vu=\dfrac{-25}{-50}=0.5$ (positive), so the image is virtual, erect, and diminished to half the size of the object, formed 25 cm from the lens on the same side as the object.
27.↑ back to the question
(a) As the wavelength of light increases, the refractive index $\mu$ of the lens material decreases slightly, so $(\mu-1)$ decreases; since $f\propto\dfrac{1}{\mu-1}$ , the focal length increases and the power $P=1/f$ decreases. (b) For a biconvex lens, $R_1=+R$ , $R_2=-R$ . For $f=R$ : $\;\dfrac1R=(\mu-1)\left(\dfrac1R+\dfrac1R\right)=(\mu-1)\dfrac2R\;\Rightarrow\;\mu-1=\dfrac12\;\Rightarrow\;\mu=\dfrac32$ (c) In air: $\;\dfrac1{f_{air}}=(\mu_g-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=(1.5-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=-\dfrac1{20}$ $\Rightarrow\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=-\dfrac1{10}$ In water: relative refractive index $\;^w\mu_g=\dfrac{\mu_g}{\mu_w}=\dfrac{1.5}{4/3}=1.125$ $\dfrac1{f_w}=(1.125-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=0.125\times(-0.1)=-0.0125$ $\Rightarrow f_w=-80\text{ cm}$ OR (c) $P=5$ D $\Rightarrow f=\dfrac1P=\dfrac{100}{5}=20$ cm (converging lens). Erect image $\Rightarrow$ virtual image, $m=+3=\dfrac vu\;\Rightarrow\;v=3u$ Using the lens formula: $\;\dfrac1v-\dfrac1u=\dfrac1f\;\Rightarrow\;\dfrac1{3u}-\dfrac1u=\dfrac1{20}\;\Rightarrow\;\dfrac{1-3}{3u}=\dfrac1{20}\;\Rightarrow\;\dfrac{-2}{3u}=\dfrac1{20}$ $\Rightarrow u=-\dfrac{40}{3}\approx-13.33$ cm, and $v=3u=-40$ cm. So the object is $\dfrac{40}{3}\approx13.33$ cm in front of the lens, and the (virtual) image forms 40 cm in front of the lens, on the same side.
28.↑ back to the question
(a) (ii): As wavelength decreases, refractive index increases, so $(\mu-1)$ increases, focal length decreases, and power $P=1/f$ increases. Hence option (ii). (b) (iv): $P=\dfrac{100}{f(\text{cm})}=\dfrac{100}{-40}=-2.5$ D. Hence option (iv). (c) (iv): In air, $\dfrac1{f_1}=(\mu_g-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\;\Rightarrow\;\dfrac1{20}=(1.5-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\qquad\text{...(1)}$ In the medium, for $f_2=60$ cm (converging, so $f_2$ positive): $\;\dfrac1{f_2}=\left(\dfrac{\mu_g}{\mu}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\;\Rightarrow\;\dfrac1{60}=\left(\dfrac{1.5}{\mu}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\qquad\text{...(2)}$ Dividing (1) by (2): $\;\dfrac{60}{20}=\dfrac{1.5-1}{1.5/\mu-1}\;\Rightarrow\;3=\dfrac{0.5}{1.5/\mu-1}\;\Rightarrow\;1.5/\mu-1=\dfrac16\;\Rightarrow\;\dfrac{1.5}{\mu}=1.1\overline6\;\Rightarrow\;\mu\approx1.2$ Hence option (iv). (d) (ii): The angle of deviation (and hence the bending) is greatest for violet light, so violet light converges nearest to the lens. Hence option (ii). (e) (iv): For $L_1$ : $u_1=-\infty$ , $f_1=15$ cm $\Rightarrow v_1=15$ cm. For the final image (from $L_2$ ) to form at infinity, this intermediate image must lie exactly at the focus of $L_2$ , i.e. the object distance for $L_2$ must be $f_2=25$ cm: $\;d-15=25\;\Rightarrow\;d=40$ cm. Hence option (iv).
29.↑ back to the question
Let the focal length of the original equi-convex lens be $f$ , and that of each plano-convex half (after cutting perpendicular to the axis, separating the two curved faces) be $F$ . Combining the two halves in contact reproduces the original lens: $\;\dfrac1f=\dfrac1F+\dfrac1F=\dfrac2F\;\Rightarrow\;F=2f$ So each half's power is $\;P=\dfrac1F=\dfrac1{2f}$ — half the power (equivalently, the reciprocal of twice the original focal length). For the double convex lens: $P=+5$ D, $\mu=1.55$ , $R_1=R_2=R$ (equal radii, opposite signs by convention: $R_1=+R$ , $R_2=-R$ ). Using the lens-maker's formula: $\;P=(\mu-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\;\Rightarrow\;5=(1.55-1)\left(\dfrac1R+\dfrac1R\right)=0.55\times\dfrac2R$ $\Rightarrow R=\dfrac{0.55\times2}{5}=0.22\text{ m}=22\text{ cm}$
30.↑ back to the question
In each case, the needle coincides with its own image only when the light retraces its path — i.e. when the object is placed exactly at the focus of the (lens + liquid-lens + mirror) combination. With the liquid present, the equivalent focal length of the (biconvex lens + liquid plano-concave lens) combination equals $x$ : so, if $f_1=y$ is the biconvex lens's own focal length (found from the second experiment, with the liquid removed) and $f_2$ is the liquid lens's focal length: $\;\dfrac1x=\dfrac1{f_1}+\dfrac1{f_2}=\dfrac1y+\dfrac1{f_2}$ $\Rightarrow\dfrac1{f_2}=\dfrac1x-\dfrac1y=\dfrac{y-x}{xy}\;\Rightarrow\;f_2=\dfrac{xy}{y-x}$ The liquid layer trapped between the lens's lower curved surface and the plane mirror forms a plano-concave lens, with $R_1=-R$ (curved face) and $R_2=\infty$ (flat face, against the mirror). Using the lens-maker's formula for this liquid lens (refractive index $\mu_l$ ): $\;\dfrac1{f_2}=(\mu_l-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=(\mu_l-1)\left(-\dfrac1R\right)$ $\Rightarrow\dfrac{y-x}{xy}=-\dfrac{(\mu_l-1)}{R}\;\Rightarrow\;\mu_l-1=\dfrac{(x-y)R}{xy}$ $\mu_l=1+\dfrac{(x-y)R}{xy}$
31.↑ back to the question
For lens $L_1$ : $u=-15$ cm, $f_{L_1}=20$ cm. Using the lens formula: $\;\dfrac1v-\dfrac1u=\dfrac1f\;\Rightarrow\;\dfrac1v=\dfrac1{20}+\dfrac1{-15}=\dfrac{3-4}{60}=-\dfrac1{60}$ $\Rightarrow v=-60$ cm — the image formed by $L_1$ is virtual, 60 cm on the same side as the object $O$ . This image now acts as the object for lens $L_2$ , which is 20 cm from $L_1$ : since the image is 60 cm on the object's side of $L_1$ , its distance from $L_2$ is $u_{L_2}=v-20=-60-20=-80$ cm (a real object for $L_2$ ). Given the final image forms 80 cm from $L_2$ : $v_{L_2}=+80$ cm. Using the lens formula for $L_2$ : $\;\dfrac1{v_{L_2}}-\dfrac1{u_{L_2}}=\dfrac1{f_{L_2}}\;\Rightarrow\;\dfrac1{80}-\dfrac1{-80}=\dfrac1{f_{L_2}}=\dfrac{1}{80}+\dfrac1{80}=\dfrac{2}{80}=\dfrac1{40}$ $\Rightarrow f_{L_2}=40$ cm.
32.↑ back to the question
(a) An object is placed at $O$ . Lens $A$ (focal length $f_1$ ) produces an image at $I_1$ , which serves as a virtual object for lens $B$ (focal length $f_2$ ), which produces the final image at $I$ . The lenses are thin, so their optical centres coincide. For lens $A$ : $\;\dfrac1{v_1}-\dfrac1u=\dfrac1{f_1}\qquad\text{...(i)}$ For lens $B$ : $\;\dfrac1v-\dfrac1{v_1}=\dfrac1{f_2}\qquad\text{...(ii)}$ Adding (i) and (ii), the $1/v_1$ terms cancel: $\;\dfrac1v-\dfrac1u=\dfrac1{f_1}+\dfrac1{f_2}\qquad\text{...(iii)}$ If the combination is equivalent to a single lens of focal length $f$ : $\;\dfrac1v-\dfrac1u=\dfrac1f\qquad\text{...(iv)}$ From (iii) and (iv): $\;\dfrac1f=\dfrac1{f_1}+\dfrac1{f_2}\;$ , and correspondingly $P=P_1+P_2$ . (b) For lens $L_1$ ( $f=+10$ cm): $u=-30$ cm. $\;\dfrac1{v_1}-\dfrac1u=\dfrac1f\;\Rightarrow\;\dfrac1{v_1}=\dfrac1{10}-\dfrac1{30}=\dfrac{3-1}{30}=\dfrac2{30}$ $\Rightarrow v_1=15$ cm. For lens $L_2$ (5 cm from $L_1$ , $f=-10$ cm): the image from $L_1$ lies $15-5=10$ cm beyond $L_2$ , acting as a virtual object, $u=+10$ cm. $\dfrac1{v_2}=\dfrac1f+\dfrac1u=\dfrac1{-10}+\dfrac1{10}=0\;\Rightarrow\;v_2=\infty$ — the rays emerge parallel after $L_2$ . For lens $L_3$ (10 cm from $L_2$ , $f=+30$ cm): since the incoming rays are parallel (object effectively at infinity), the final image forms at the focus of $L_3$ . Hence the final image is formed at a distance of 30 cm from $L_3$ (the last lens).
33.↑ back to the question
Consider two thin convex lenses $A$ and $B$ , of focal lengths $f_1$ and $f_2$ , placed in contact. For an object at $O$ , lens $A$ forms an intermediate image at $I'$ (distance $v'$ ), which acts as the object for lens $B$ , which forms the final image at $I$ (distance $v$ ). For lens $A$ : $\;\dfrac1{v'}-\dfrac1u=\dfrac1{f_1}\qquad\text{...(i)}$ For lens $B$ : $\;\dfrac1v-\dfrac1{v'}=\dfrac1{f_2}\qquad\text{...(ii)}$ Adding (i) and (ii): $\;\dfrac1v-\dfrac1u=\dfrac1{f_1}+\dfrac1{f_2}$ Since the combination behaves as a single lens of focal length $f$ with $\dfrac1v-\dfrac1u=\dfrac1f$ , we get: $\;\dfrac1f=\dfrac1{f_1}+\dfrac1{f_2}$ Since power $P=1/f$ : $\;P_T=P_1+P_2$
34.Answer: (A)↑ back to the question
Since the image is of the same size as the object (the window) and is real (formed on a wall/screen), the object must be placed at $2F$ from the lens, with the image also forming at $2F$ on the other side — a standard unit-magnification real-image configuration for a converging lens. So the object-to-image distance $d=2F+2F=4F\;\Rightarrow\;F=\dfrac d4$ , and since this is a real image via a converging lens, $F$ is positive. Hence option (a): $f=+\dfrac d4$ .
35.Answer: (A)↑ back to the question
A lens becomes invisible (behaves as a plane plate with no net bending of light) when the refractive index of the surrounding liquid exactly equals the refractive index of the lens material, since then $(\mu_{lens}/\mu_{liquid}-1)=0$ and the lens-maker's formula gives $1/f=0$ . So the refractive index of the liquid must be 1.47. Hence option (a).
36.Answer: (B)↑ back to the question
For the original bi-concave lens: $\;\dfrac1F=(\mu-1)\left(\dfrac1{-R}-\dfrac1R\right)=-\dfrac{2(\mu-1)}{R}$ For each plano-concave half (one curved face of radius $R$ , one flat face): $\;\dfrac1{F'}=(\mu-1)\left(\dfrac1{-R}-\dfrac1\infty\right)=-\dfrac{\mu-1}{R}$ Comparing, $\;\dfrac1{F'}=\dfrac12\times\dfrac1F\;\Rightarrow\;F'=2F\;\Rightarrow\;P'=\dfrac{P}{2}$ Hence option (b).