Optical Instruments

Ray Optics and Optical Instruments · CBSE

Theory

What this subtopic is examined on

Simple microscope: magnifying power at the near point and at infinity.
Compound microscope: ray diagram, derivation of the magnifying power in both adjustments, why both focal lengths must be short, numericals on separation and magnification.
Astronomical refracting telescope: ray diagram, magnifying power, tube length, near-point adjustment, height of the image formed by the objective.
Reflecting (Cassegrain) telescope: schematic, and its advantages over a refracting telescope.
Lens-selection questions given powers and apertures, and comparison of a microscope with a telescope.

Simple microscope (magnifying glass)

Diagram in words

A converging lens of short focal length is held close to the eye, with the object placed within one focal length of it. Two rays from the top of the object — one parallel to the axis emerging through the second focus, one through the optical centre undeviated — diverge after refraction; produced backwards they meet at the top of an erect, virtual, magnified image on the same side as the object, at the near point (or, with the object exactly at $f$ , at infinity).

Results

Image at the near point $D = 25$ cm. From $m = \dfrac{v}{u}$ and the lens formula with $v = -D$ :

$m = 1 + \dfrac{D}{f}$

Image at infinity (object at the focus, relaxed eye). The angle subtended by the object at the near point without the lens is $\theta_o = h/D$ ; with the lens the image subtends $\theta_i = h/f$ . So the angular magnification is

$m = \dfrac{\theta_i}{\theta_o} = \dfrac{D}{f}$

The infinity setting gives one less than the near-point value but is more comfortable, since the eye is relaxed. In later work on microscopes and telescopes the image is assumed to be at infinity unless stated otherwise.
What a magnifier actually does: the angular size of the object equals the angular size of the image, but the lens lets the object be brought much closer than $D$ and still be seen clearly, so it subtends a bigger angle than it could unaided.
A lens of $f = 5$ cm gives $m = 1 + 25/5 = 6$ at the near point and $25/5 = 5$ at infinity. A lens of $f = 30$ cm gives $m = 25/30 < 1$ at infinity and so is useless as a magnifier.
The maximum magnification of a realistic simple microscope is only about 9; larger values need impractically short focal lengths, whose strong curvature brings severe aberrations. That is why the compound microscope exists.

Compound microscope

Diagram in words

Two converging lenses coaxial in a tube: the objective, of very short focal length $f_o$ and small aperture, near the object; the eyepiece, of short focal length $f_e$ and slightly larger aperture, near the eye.
The tiny object $AB$ is placed just beyond the first focus of the objective. The objective forms a real, inverted, magnified image $A'B'$ inside the tube.
$A'B'$ lies at (for the infinity setting) or just within (for the near-point setting) the first focus of the eyepiece. The eyepiece acts as a simple magnifier on it and forms the final image $A''B''$ , virtual, further magnified, and inverted with respect to the original object — at infinity, or at the near point $D$ .
Mark $f_o$ , $f_e$ and the tube length $L$ , the distance between the second focal point of the objective and the first focal point of the eyepiece.

Derivation

Magnification of the objective:

$m_o = \dfrac{h'}{h} = \dfrac{v_o}{u_o} \approx \dfrac{L}{f_o}$

The approximation uses $v_o \approx L$ (the first image forms near the far end of the tube) and $u_o \approx f_o$ (the object sits just outside the focus).
Magnification of the eyepiece, working as a simple magnifier:

$m_e = 1 + \dfrac{D}{f_e} \quad \text{(final image at the near point)}, \qquad m_e = \dfrac{D}{f_e} \quad \text{(final image at infinity)}$

Since the image formed by the first lens is the object for the second, the magnifications multiply:

$m = m_o\times m_e = \dfrac{v_o}{u_o}\left(1+\dfrac{D}{f_e}\right) \approx \dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$

$m = \dfrac{L}{f_o}\times\dfrac{D}{f_e} \quad \text{(normal adjustment, final image at infinity)}$

Why both focal lengths must be short: $m$ is inversely proportional to both $f_o$ and $f_e$ , so shortening either raises the magnification. In practice it is hard to make $f$ much smaller than 1 cm, and a large $L$ needs large lenses.
Definition to quote for the magnifying power at infinity: the ratio of the angle subtended at the eye by the final image to the angle subtended at the unaided eye by the object placed at the least distance of distinct vision.
Modern microscopes use multi-component lenses for both objective and eyepiece to minimise chromatic and spherical aberration and improve image quality. Illumination and contrast also matter to the visibility of the image.
With $f_o = 1.0$ cm, $f_e = 2.0$ cm and $L = 20$ cm, $m = \dfrac{20}{1}\times\dfrac{25}{2} = 250$ .

Astronomical (refracting) telescope

Diagram in words

Two converging lenses coaxial in a tube: the objective, of large focal length $f_o$ and large aperture, facing the distant object; the eyepiece, of short focal length $f_e$ , at the eye end.
Rays from a distant object arrive as a parallel bundle inclined at a small angle $\alpha$ to the axis. The objective brings them to a real, inverted, diminished image $A'B'$ at its second focal point.
In normal adjustment the eyepiece is placed so that $A'B'$ lies at its first focal point; the emergent rays are then parallel again, and the final image is virtual, inverted and at infinity, subtending angle $\beta$ at the eye.
Mark $\alpha$ and $\beta$ , and the tube length $f_o + f_e$ .

Derivation

With $h$ the height of the intermediate image, and for small angles,

$\alpha \approx \dfrac{h}{f_o}, \qquad \beta \approx \dfrac{h}{f_e}$

$m = \dfrac{\beta}{\alpha} = \dfrac{h/f_e}{h/f_o} = \dfrac{f_o}{f_e}$

Length of the telescope tube in normal adjustment: $L = f_o + f_e$ .
Definition to quote: the magnifying power of a telescope is the ratio of the angle subtended at the eye by the final image to the angle subtended at the unaided eye by the object.
Final image at the least distance of distinct vision (near-point adjustment): the eyepiece must be moved slightly closer to the intermediate image than $f_e$ , and

$m = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right)$

Height (or diameter) of the intermediate image of a distant object subtending angle $\theta$ :

$h = f_o\,\theta = f_o\times\dfrac{\text{size of the object}}{\text{distance of the object}}$

To increase the magnifying power: increase $f_o$ and decrease $f_e$ .
Why a large objective aperture: it gathers more light, so faint distant objects give a brighter image, and it raises the resolving power — the ability to see two objects in nearly the same direction as separate.
Terrestrial telescopes add a pair of inverting lenses so the final image is erect.

Reflecting telescope (Cassegrain)

Diagram in words

A large concave paraboloidal primary mirror at the closed end of the tube, with a small hole at its centre. Parallel rays from a distant object fall on it and are reflected towards its focus.
Before they converge, a small convex secondary mirror mounted on the axis intercepts them and reflects them back through the hole in the primary.
The image forms just behind the hole and is viewed through an eyepiece placed there.

Why mirrors are preferred for large telescopes

There is no chromatic aberration at all in a mirror, because reflection does not depend on the colour of the light; a lens objective refracts and therefore focuses different colours at slightly different points.
Using a paraboloidal rather than spherical mirror removes spherical aberration too.
A mirror can be supported over its entire back surface, whereas a lens can only be held at its rim and sags under its own weight. Large mirrors are therefore mechanically practical, while a large lens is not — the biggest lens objective in use is 40 inch (about 1.02 m) at Yerkes, whereas the Keck reflectors are 10 m across.
A mirror weighs much less than a lens of the same optical quality, is cheaper to make free of defects, and absorbs far less light, so the image is brighter.
Larger apertures then mean greater light-gathering power and higher resolving power.
Examples: the 200 inch (about 5.08 m) Mt Palomar telescope; the 2.34 m Cassegrain at Kavalur, Tamil Nadu, the largest in India; the pair of 10 m Keck telescopes in Hawaii.
The one drawback of a reflector is that the objective mirror focuses the light inside the tube, so the eyepiece or observer cage obstructs some of the incoming light; the Cassegrain arrangement solves this and also gives a long focal length in a short tube.

Choosing lenses in an exam question

Requirement	Microscope	Telescope
Objective focal length	Smallest of the set	Largest of the set
Objective aperture	Small	Large
Eyepiece focal length	Short, but larger than $f_o$	Shortest of the set

Given $L_1$ (3 D, aperture 8 cm), $L_2$ (6 D, aperture 1 cm) and $L_3$ (10 D, aperture 1 cm), the focal lengths are 33.3 cm, 16.7 cm and 10 cm. For a microscope use $L_3$ as objective and $L_2$ as eyepiece; for a telescope use $L_1$ as objective (largest $f$ , largest aperture) and $L_3$ as eyepiece.
Given 0.5 D, 4 D and 10 D lenses ( $f = 200$ cm, 25 cm, 10 cm) for a telescope: the 0.5 D lens is the objective and the 10 D lens the eyepiece, giving the largest $m = f_o/f_e$ .
Identifying the instrument from data: if both focal lengths are short and a finite tube length is quoted, it is a compound microscope; if the objective focal length is far larger than the eyepiece's, it is a telescope. Simply interchanging the two lenses of a microscope does not make a usable telescope, because a telescope objective needs large $f$ and large aperture.

Comparison table

Feature	Compound microscope	Astronomical telescope
Object	Tiny, close by	Large, very distant
Objective	Short $f$ , small aperture	Long $f$ , large aperture
Eyepiece relative to objective	$f_e > f_o$	$f_e \ll f_o$
Intermediate image	Real, inverted, magnified	Real, inverted, diminished
Final image	Virtual, inverted, magnified	Virtual, inverted
Magnifying power (infinity)	$\dfrac{L}{f_o}\times\dfrac{D}{f_e}$	$\dfrac{f_o}{f_e}$
Tube length (infinity)	$v_o + f_e$	$f_o + f_e$

Numerical patterns CBSE repeats

Separation in a compound microscope, final image at the near point. Find $m_e = 1 + D/f_e$ , hence $m_o = m/m_e$ ; apply the lens formula to the objective with $v_o = -m_o u_o$ to get $u_o$ and $v_o$ ; apply it to the eyepiece with $v_e = -D$ to get $u_e$ ; the separation is $v_o + \lvert u_e\rvert$ .
Worked instance: $f_o = 1.25$ cm, $f_e = 5$ cm, required $m = 30$ at the near point. Then $m_e = 6$ , $m_o = 5$ , $u_o = -1.5$ cm, $v_o = 7.5$ cm, $u_e = -25/6 \approx -4.17$ cm, and the separation is $7.5 + 4.17 = 11.67$ cm.
Final image at infinity instead: $f_o = 1.25$ cm, object at 2.5 cm gives $v_o = 2.5$ cm; the intermediate image must sit at the focus of the eyepiece, so the separation is $v_o + f_e = 2.5 + 5 = 7.5$ cm.
Objective of power 100 D and eyepiece of 40 D with $L = 20$ cm: $f_o = 1$ cm, $f_e = 2.5$ cm, so it is a compound microscope and $m = \dfrac{20}{1}\times\dfrac{25}{2.5} = 200$ .
Compound microscope with $u_o = -1.5$ cm, $f_o = 1.25$ cm, $f_e = 5$ cm, image at the near point: $v_o = 7.5$ cm, $m_o = -5$ , $m_e = 6$ , so $m = -30$ , of magnitude 30.
Given total magnification 20, eyepiece magnification 5, separation 14 cm and $D = 20$ cm: $m_e = 1 + D/f_e = 5$ gives $f_e = 5$ cm; $m_o = 4$ ; and $m = \dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$ with $L = 14$ cm gives $f_o = 3.5$ cm.
Telescope, $f_o = 150$ cm, $f_e = 6$ cm at infinity: $m = 25$ and tube length $156$ cm.
Telescope, $f_o = 140$ cm, $f_e = 5$ cm: normal adjustment $m = 28$ ; near point $m = 28\left(1+\dfrac{5}{25}\right) = 33.6$ .
Telescope in normal adjustment with $m = 2.9$ and separation 150 cm: $f_o = 2.9f_e$ and $f_o + f_e = 150$ , so $f_e \approx 38.5$ cm and $f_o \approx 111.5$ cm.
Objective image of a distant object: $f_o = 150$ cm viewing a 100 m tower 3 km away gives $h = 1.5\times\dfrac{100}{3000} = 0.05$ m $= 5$ cm, real and inverted.
The Moon through a giant telescope: with $f_o = 15$ m, Moon diameter $3.48\times10^6$ m and distance $3.8\times10^8$ m, the image diameter is $\dfrac{3.48\times10^6\times15}{3.8\times10^8} \approx 0.137$ m $= 13.7$ cm. With $f_o = 20$ m and Moon diameter $3.5\times10^6$ m, it is about $18.4$ cm.
Simple magnifier numerical: object 3.0 cm from a lens of $f = 4.0$ cm gives $v = -12$ cm (virtual, same side) and $m = v/u = 4$ .

Standard results

Simple microscope: $m = 1 + D/f$ (near point), $m = D/f$ (infinity), $D = 25$ cm.
Compound microscope: $m = \dfrac{v_o}{u_o}\left(1+\dfrac{D}{f_e}\right)$ at the near point, $\dfrac{L}{f_o}\times\dfrac{D}{f_e}$ at infinity; both focal lengths short; final image virtual and inverted.
Telescope: $m = \dfrac{f_o}{f_e}$ and $L = f_o+f_e$ at infinity, $m = \dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)$ at the near point; large $f_o$ and large aperture, short $f_e$ .
Aperture governs brightness and resolving power, never magnification.
The image at infinity is preferred because the eye is then fully relaxed and viewing causes least strain.
A reflecting telescope beats a refractor on chromatic aberration, achievable aperture, brightness, weight and cost.

Theory Questions Asked on This Subtopic

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

1
In a simple microscope, a convex lens of focal length 5 cm is used. Calculate the magnifying power when the object is placed at the focus of the lens.
1 markCBSE Class 12 Physics — Very Short Answer · 2021 · Section A, 1 markAnswer 1 ↓
2
A compound microscope is used because a realistic simple microscope does not have ____ magnification.
1 markCBSE Class 12 Physics — Multiple Choice · 2020, R · Section A, 1 markAnswer 2 ↓
3
Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain.
2 marksCBSE Class 12 Physics — SA · Delhi 2017 · Section B, 2 marksAnswer 3 ↓
4
You are given two converging lenses of focal lengths 1.25 cm and 5 cm to design a compound microscope. If it is desired to have a magnification of 30 (final image at the near point), find the separation between the objective and the eyepiece.
2 marksCBSE Class 12 Physics — SA · AI 2015, Ev · Section B, 2 marksAnswer 4 ↓
5
With the help of a ray diagram, show how a compound microscope forms a magnified image of a tiny object at the least distance of distinct vision. Hence derive an expression for the magnification produced by it. OR With the help of a ray diagram, explain the formation of the image in a compound microscope when the final image is formed at the near point. Obtain the expression for the magnifying power in this case.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marksAnswer 5 ↓
6
(i) Draw a labelled ray diagram showing the formation of the image at the least distance of distinct vision by a simple microscope (magnifier). (ii) A small object is placed at a distance of 3.0 cm from a magnifier of focal length 4.0 cm. Find (I) the position of the image formed, (II) the linear magnification produced.
3 marksCBSE Class 12 Physics — SA · Section C, 3 marksAnswer 6 ↓
7
An optical instrument uses an objective lens of power 100 D and an eyepiece of power 40 D. The final image is formed at infinity when the tube length of the instrument is kept at 20 cm. (a) Identify the optical instrument. (b) Calculate the angular magnification produced by the instrument.
3 marksCBSE Class 12 Physics — SA · 2020, Ev · Section C, 3 marksAnswer 7 ↓
8
(a) Draw a ray diagram for the formation of an image by a compound microscope. (b) You are given the following three lenses. Which two lenses will you use as the eyepiece and as the objective to construct a compound microscope? Give reason. Lens $L_1$ : power 3D, aperture 8 cm. Lens $L_2$ : power 6D, aperture 1 cm. Lens $L_3$ : power 10D, aperture 1 cm.
5 marksCBSE Class 12 Physics — Long Answer · AI 2017 · Section E, 5 marksAnswer 8 ↓
9
Draw a ray diagram showing the image formation by a compound microscope. Obtain the expression for the total magnification produced, when the final image is formed at infinity (normal adjustment).
5 marksCBSE Class 12 Physics — Long Answer · AI 2015C · Section E, 5 marksAnswer 9 ↓
10
(a) Draw a labelled ray diagram showing the formation of the final image by a compound microscope at the least distance of distinct vision. (b) The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece is 5. The microscope is focussed on a certain object. The distance between the objective and the eyepiece is observed to be 14 cm. If the least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eyepiece.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2014, An · Section E, 5 marksAnswer 10 ↓
11
(i) Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification when the final image is formed at the near point. (ii) In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eyepiece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope.
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 11 ↓
12
A compound microscope consists of two converging lenses. One of them, of smaller aperture and smaller focal length, is called the objective, and the other, of slightly larger aperture and slightly larger focal length, is called the eyepiece. Both lenses are fitted in a tube with an arrangement to vary the distance between them. A tiny object is placed in front of the objective at a distance slightly greater than its focal length. The objective produces an image of the object which acts as an object for the eyepiece, which in turn produces the final magnified image. (I) In a compound microscope, the images formed by the objective and the eyepiece are respectively: (a) virtual, real (b) real, virtual (c) virtual, virtual (d) real, real (II) The magnification due to a compound microscope does NOT depend upon: (a) the aperture of the objective and the eyepiece (b) the focal length of the objective and the eyepiece (c) the length of the tube (d) the colour of light used (III) Which of the following is NOT correct in the context of a compound microscope? (a) Both the lenses are of short focal lengths. (b) The magnifying power increases by decreasing the focal lengths of the two lenses. (c) The distance between the two lenses is more than $(f_o+f_e)$ . (d) The microscope can be used as a telescope by interchanging the two lenses. (IV) A compound microscope consists of an objective of magnifying power 10X and an eyepiece of magnifying power 20X. The magnification due to the microscope would be: (a) 2 (b) 10 (c) 30 (d) 200 (V) The focal lengths of the objective and eyepiece of a compound microscope are 1.2 cm and 3.0 cm respectively. The object is placed at a distance of 1.25 cm from the objective. If the final image is formed at infinity, the magnifying power of the microscope would be: (a) 100 (b) 150 (c) 200 (d) 250
5 marksCBSE Class 12 Physics — Case · AI 2017, An · Section E, 5 marksAnswer 12 ↓
13
Why should the objective of a telescope have a large focal length and a large aperture? Justify your answer.
2 marksCBSE Class 12 Physics — SA · Delhi 2017, U · Section B, 2 marksAnswer 13 ↓
14
(a) Draw a ray diagram showing the formation of an image by a reflecting telescope. (b) Write two advantages of a reflecting telescope over a refracting telescope. OR Draw a schematic arrangement of a reflecting (Cassegrain) telescope showing how rays coming from a distant object are received at the eyepiece. Write two important advantages of this arrangement over a refracting telescope.
5 marksCBSE Class 12 Physics — Long Answer · AI 2017 · Section E, 5 marksAnswer 14 ↓
15
A small telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
2 marksCBSE Class 12 Physics — SA · AI 2015 · Section B, 2 marksAnswer 15 ↓
16
With the help of a ray diagram, explain the working of a reflecting telescope. Mention two advantages of a reflecting telescope over a refracting telescope.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marksAnswer 16 ↓
17
(a)(i) Draw a labelled ray diagram showing the formation of the image at infinity by an astronomical telescope. (ii) A telescope consists of an objective of focal length 150 cm and an eyepiece of focal length 6.0 cm. If the final image is formed at infinity, calculate (I) the length of the tube in this adjustment, and (II) the magnification produced.
4 marksCBSE Class 12 Physics — Long Answer · Term II 2021-22 · Section E, 4 marksAnswer 17 ↓
18
What is the difference in the construction of an astronomical telescope and a compound microscope? The focal lengths of the objective and eyepiece of a compound microscope are 1.25 cm and 5.0 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30, when the final image is formed at the near point.
3 marksCBSE Class 12 Physics — SA · 2020 · Section C, 3 marksAnswer 18 ↓
19
Draw a labelled ray diagram of an astronomical telescope in the near-point-adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is $3.48\times10^6$ m and the radius of the lunar orbit (Earth-Moon distance) is $3.8\times10^8$ m.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2019, Ev · Section E, 5 marksAnswer 19 ↓
20
Draw a ray diagram to show the image formation of a distant object by a refracting telescope. Write the expression for its angular magnification in terms of the focal lengths of the lenses used. State the important considerations required to achieve a large resolution, and their consequent limitations.
5 marksCBSE Class 12 Physics — Long Answer · AI 2019, Ev · Section E, 5 marksAnswer 20 ↓
21
(a) Draw a ray diagram depicting the formation of the image by an astronomical telescope in normal adjustment. (b) You are given the following three lenses. Which two lenses will you use as the eyepiece and as the objective to construct an astronomical telescope? Give reason. Lens $L_1$ : power 3D, aperture 8 cm. Lens $L_2$ : power 6D, aperture 1 cm. Lens $L_3$ : power 10D, aperture 1 cm.
4 marksCBSE Class 12 Physics — Long Answer · AI 2016, Ev · Section E, 4 marksAnswer 21 ↓
22
Which two of the following lenses $L_1$ , $L_2$ , $L_3$ will you select as the objective and eyepiece for constructing the best possible (i) telescope, (ii) microscope? Give reasons. Lens $L_1$ : power 6D, aperture 1 cm. Lens $L_2$ : power 3D, aperture 8 cm. Lens $L_3$ : power 10D, aperture 1 cm.
3 marksCBSE Class 12 Physics — SA · Delhi 2015C · Section C, 3 marksAnswer 22 ↓
23
(i) Draw a ray diagram to show how the final image is formed at infinity in an astronomical refracting telescope. Obtain an expression for its magnifying power. (ii) Two thin lenses $L_1$ and $L_2$ — $L_1$ a convex lens of focal length 24 cm, and $L_2$ a concave lens of focal length 18 cm — are placed coaxially at a separation of 45 cm. A 1 cm tall object is placed in front of lens $L_1$ at a distance of 36 cm. Find the location and height of the image formed by the combination.
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 23 ↓
24
(i) Draw a ray diagram for the formation of the image of an object by an astronomical telescope in normal adjustment. Obtain the expression for its magnifying power. (ii) The magnifying power of an astronomical telescope in normal adjustment is 2.9, and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 24 ↓
25
An astronomical telescope has an objective lens of focal length 20 m and an eyepiece of focal length 1 cm. (a) Find the angular magnification of the telescope. (b) If the telescope is used to view the Moon, find the diameter of the image formed by the objective lens. The diameter of the Moon is $3.5\times10^6$ m and the radius of the lunar orbit is $3.8\times10^8$ m.
3 marksCBSE Class 12 Physics — SA · 2020 · Section C, 3 marksAnswer 25 ↓
26
Explain two advantages of a reflecting telescope over a refracting telescope.
2 marksCBSE Class 12 Physics — SA · 2018, U · Section B, 2 marksAnswer 26 ↓
27
Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment. Define its magnifying power. You are given three lenses of power 0.5D, 4D and 10D to design a telescope. (i) Which lenses should be used as the objective and the eyepiece? Justify your answer. (ii) Why is a large aperture preferred for the objective?
5 marksCBSE Class 12 Physics — Long Answer · AI 2016, Ev · Section E, 5 marksAnswer 27 ↓
28
(a) Draw a labelled schematic ray diagram of an astronomical telescope in normal adjustment. (b) Which two aberrations do the objectives of refracting telescopes suffer from? How are these overcome in a reflecting telescope?
5 marksCBSE Class 12 Physics — Long Answer · Foreign 2016 · Section E, 5 marksAnswer 28 ↓
29
Draw a ray diagram showing the image formation by a refracting telescope. Define its magnifying power and write the two important factors considered to increase the magnifying power. Describe briefly the two main limitations, and explain how these can be minimized in a reflecting telescope.
5 marksCBSE Class 12 Physics — Long Answer · Foreign 2015 · Section E, 5 marksAnswer 29 ↓
30
(a) Draw a labelled ray diagram of an astronomical telescope showing the image formation of a distant object. State the main considerations in selecting the objective and eyepiece lenses in order to have a large magnifying power and high resolution. (b) A compound microscope has an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. A small object is kept at 2.5 cm from the objective. If the final image is formed at infinity, find the distance between the objective and the eyepiece.
5 marksCBSE Class 12 Physics — Long Answer · Section E, 5 marksAnswer 30 ↓

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
When the object is at the focus, the image forms at infinity (normal adjustment): $\;M=\dfrac Df=\dfrac{25}5=5$
2.↑ back to the question
Answer: (sufficiently) large magnification. A simple microscope of very short focal length would be needed to get large magnification, but such small, strongly-curved lenses suffer from severe aberrations; hence a single (simple) microscope cannot provide sufficiently large / adequate magnification for practical use — this need is met instead by a compound microscope.
3.↑ back to the question
Magnifying power (final image at infinity, normal adjustment) is the ratio of the angle subtended at the eye by the final image to the angle subtended at the (unaided) eye by the object when placed at the least distance of distinct vision: $\;M=\dfrac L{f_o}\times\dfrac D{f_e}$ , where $L$ is the tube length (separation between the lenses). Since $M$ is inversely proportional to both $f_o$ and $f_e$ , decreasing either focal length increases the overall magnifying power. Hence both the objective and the eyepiece are made with short focal lengths, so that the compound microscope can achieve a large overall magnification.
4.↑ back to the question
Here $f_o=1.25$ cm (short focal length $\Rightarrow$ objective) and $f_e=5$ cm (eyepiece). For the final image at the near point ( $D=25$ cm): $\;m_e=1+\dfrac D{f_e}=1+\dfrac{25}5=6$ Overall magnification $m=m_o\times m_e=30\;\Rightarrow\;m_o=\dfrac{30}6=5$ For the objective, with $|v_o|=5|u_o|$ (real, inverted image of magnitude 5) and $\dfrac1{v_o}-\dfrac1{u_o}=\dfrac1{f_o}$ : taking $v_o=-5u_o$ , $\;\dfrac1{-5u_o}-\dfrac1{u_o}=\dfrac1{1.25}\;\Rightarrow\;\dfrac{-6}{5u_o}=0.8\;\Rightarrow\;u_o=-1.5\text{ cm},\ v_o=7.5\text{ cm}$ For the eyepiece, the image at the near point means $v_e=-25$ cm: $\;\dfrac1{v_e}-\dfrac1{u_e}=\dfrac1{f_e}\;\Rightarrow\;\dfrac1{-25}-\dfrac1{u_e}=\dfrac15\;\Rightarrow\;\dfrac1{u_e}=-\dfrac15-\dfrac1{25}=-\dfrac6{25}\;\Rightarrow\;u_e=-\dfrac{25}6\approx-4.17\text{ cm}$ Separation between the lenses $=v_o+|u_e|=7.5+\dfrac{25}6\approx7.5+4.17=11.67$ cm
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Ray diagram: the tiny object $O$ is placed just beyond the focus of the objective lens (short focal length, small aperture); the objective forms a real, inverted, magnified image $A'B'$ inside the tube. This image acts as the object for the eyepiece (also short focal length), which is positioned so that $A'B'$ lies between its focus and optical centre; the eyepiece then forms a final virtual, magnified image $A''B''$ at the near point (or, in normal adjustment, at infinity). Magnification produced by the objective: $\;m_o=\dfrac{v_o}{u_o}$ Magnification produced by the eyepiece (acting as a simple magnifier, final image at near point $D$ ): $\;m_e=1+\dfrac D{f_e}$ Total magnification: $\;m=m_o\times m_e=\dfrac{v_o}{u_o}\left(1+\dfrac D{f_e}\right)$ Since $u_o\approx f_o$ (object placed very close to the focus) and $v_o\approx L$ (the tube length, i.e. separation between the lenses), this is often approximated as $\;m\approx\dfrac L{f_o}\left(1+\dfrac D{f_e}\right)$
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(i) A convex lens of short focal length is held close to the eye; the object, placed within the focal length, forms a magnified, virtual, erect image on the same side as the object, at (or beyond) the near point. (ii) $u=-3.0$ cm, $f=4.0$ cm: $\;\dfrac1v-\dfrac1u=\dfrac1f\;\Rightarrow\;\dfrac1v=\dfrac14+\dfrac1{-3}=\dfrac{3-4}{12}=-\dfrac1{12}$ $\Rightarrow\;v=-12$ cm — the image is virtual, formed 12 cm from the lens, on the same side as the object. (II) Linear magnification: $\;m=\dfrac vu=\dfrac{-12}{-3}=4$
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$f_o=\dfrac1{P_o}=\dfrac1{100}\text{ m}=1\text{ cm}$ ; $\;f_e=\dfrac1{P_e}=\dfrac1{40}\text{ m}=2.5\text{ cm}$ (a) Since $f_o<f_e$ (both fairly short focal lengths) and a finite tube length is specified, this is a COMPOUND MICROSCOPE (a telescope objective would instead need a much larger focal length than the eyepiece). (b) $\;M=\dfrac L{f_o}\times\dfrac D{f_e}=\dfrac{20}1\times\dfrac{25}{2.5}=20\times10=200$
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(a) Same ray diagram as in Q74: objective forms a real, magnified, inverted intermediate image; eyepiece further magnifies this to give the final virtual image. (b) Focal lengths: $f_1=\dfrac{100}3\approx33.3$ cm, $f_2=\dfrac{100}6\approx16.7$ cm, $f_3=\dfrac{100}{10}=10$ cm. For a microscope, BOTH lenses need short focal lengths, and the objective must have the smaller focal length as well as a small aperture (to keep it close to the tiny object). $L_3$ ( $f=10$ cm, aperture 1 cm) has the shortest focal length and a small aperture, so it should be used as the OBJECTIVE. $L_2$ ( $f=16.7$ cm, aperture 1 cm) has the next-shortest focal length, so it should be used as the EYEPIECE. ( $L_1$ , with its large 8 cm aperture and longest focal length, is unsuitable for either role in a microscope.)
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Ray diagram: as in Q74, with the eyepiece now positioned so that the intermediate real image $A'B'$ falls exactly at its focus, so the final image forms at infinity. Magnification by objective: $\;m_o=\dfrac{v_o}{u_o}\approx\dfrac L{f_o}$ (since $v_o\approx L$ , the tube length, and $u_o\approx f_o$ ). Magnification by eyepiece (final image at infinity): $\;m_e=\dfrac D{f_e}$ Total magnification: $\;M=m_o\times m_e=\dfrac L{f_o}\times\dfrac D{f_e}$
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(a) Same ray diagram as Q74/Q78 (final image at the near point, $D=20$ cm here). (b) $\;m_e=1+\dfrac D{f_e}=5\;\Rightarrow\;\dfrac{20}{f_e}=4\;\Rightarrow\;f_e=5\text{ cm}$ $\;M=m_o\times m_e=20\;\Rightarrow\;m_o=\dfrac{20}5=4$ Using the approximate relation $M=-\dfrac L{f_o}\left(1+\dfrac D{f_e}\right)$ with $L=14$ cm: $\;-20=-\dfrac{14}{f_o}\times5\;\Rightarrow\;f_o=\dfrac{14\times5}{20}=3.5\text{ cm}$
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(i) Same ray diagram and derivation as Q74: $\;M=\dfrac{v_o}{u_o}\left(1+\dfrac D{f_e}\right)$ (ii) $u_o=-1.5$ cm, $f_o=1.25$ cm: $\;\dfrac1{v_o}=\dfrac1{f_o}+\dfrac1{u_o}=\dfrac1{1.25}-\dfrac1{1.5}=0.8-0.667=0.133\;\Rightarrow\;v_o=7.5\text{ cm}$ $\;m_o=\dfrac{v_o}{u_o}=\dfrac{7.5}{-1.5}=-5$ $\;m_e=1+\dfrac D{f_e}=1+\dfrac{25}5=6$ $\;M=m_o\times m_e=(-5)(6)=-30$ (magnitude 30)
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(I) (b): The objective forms a real (inverted, magnified) image of the tiny object; this real image then acts as the object for the eyepiece, which (behaving as a simple magnifier) forms a virtual final image. (II) (a): The aperture of the lenses affects the brightness and resolving power of the image, not the magnification, which depends on the focal lengths of the lenses, the tube length, and (via dispersion) the colour of light. Hence option (a). (III) (d): A telescope objective needs a LARGE focal length and LARGE aperture (opposite of a microscope objective, which needs a small focal length and small aperture), so simply interchanging the two lenses of a microscope does not turn it into a usable telescope. Hence (d) is the incorrect statement. (IV) (d): $M=m_o\times m_e=10\times20=200$ (V) (c): $u_o=-1.25$ cm, $f_o=1.2$ cm: $\;\dfrac1{v_o}=\dfrac1{1.2}-\dfrac1{1.25}=0.833-0.8=0.0333\;\Rightarrow\;v_o=30\text{ cm}$ $\;m_o=\dfrac{v_o}{u_o}=\dfrac{30}{-1.25}=-24$ ; for final image at infinity, $\;m_e=\dfrac D{f_e}=\dfrac{25}3\approx8.33$ $\;M=|m_o|\times m_e\approx24\times8.33\approx200$ . Hence option (c).
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Magnifying power of a telescope (normal adjustment) is $\;M=\dfrac{f_o}{f_e}$ , so a larger objective focal length $f_o$ directly gives a larger magnifying power. A large aperture increases the light-gathering power of the objective, producing a brighter image of distant (often faint) objects, and it also improves the resolving power of the telescope (the ability to distinguish two closely-spaced objects), since resolving power increases with aperture.
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Ray diagram (Cassegrain reflecting telescope): parallel rays from a distant object fall on a large concave (paraboloidal) primary mirror, which reflects and converges them towards a smaller convex secondary mirror placed in their path. The secondary mirror reflects the converging beam back through a small hole at the centre of the primary mirror, forming the final image which is viewed/magnified through an eyepiece placed behind the hole. Advantages of a reflecting telescope over a refracting telescope: (i) it is completely free of chromatic aberration, since mirrors reflect all colours alike (no refraction is involved); (ii) a mirror of a given aperture is far easier to support (it can be supported over its entire back surface, unlike a lens which can only be supported at its edges), so much larger apertures are achievable, giving greater light-gathering power and higher resolving power.
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Magnifying power (normal adjustment): $\;M=\dfrac{f_o}{f_e}=\dfrac{150}5=30$ Since the tower is very far away compared to $f_o$ , its image by the objective forms essentially at the focus, and (for the small angle subtended) $\;\dfrac{h_i}{f_o}\approx\dfrac{h_o}{u_o}$ $\;h_i=f_o\times\dfrac{h_o}{u_o}=1.5\text{ m}\times\dfrac{100\text{ m}}{3000\text{ m}}=1.5\times0.0333=0.05\text{ m}=5\text{ cm}$ The image formed by the objective is real and inverted, of height 5 cm.
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Same ray diagram and explanation as Q83: a large concave primary mirror converges the incoming parallel rays onto a convex secondary mirror, which reflects them out through a hole in the primary to the eyepiece. Advantages: (i) free from chromatic aberration (no refraction involved); (ii) much larger apertures are practically achievable for mirrors than for lenses (a mirror can be fully supported from behind, unlike a lens), giving greater light-gathering power and higher resolving power.
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(i) Parallel rays from a distant object are brought to focus by the objective at its focal point, which coincides with the focal point of the eyepiece (normal adjustment); the eyepiece then renders these rays parallel again, forming a final, inverted, virtual image at infinity. (I) Tube length: $\;L=f_o+f_e=150+6=156\text{ cm}$ (II) Magnification: $\;M=\dfrac{f_o}{f_e}=\dfrac{150}6=25$
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Difference: in an astronomical telescope, the objective has a LARGE focal length and LARGE aperture (to view distant objects with good resolution), while the eyepiece has a short focal length. In a compound microscope, BOTH the objective and eyepiece have short focal lengths, with the objective's aperture and focal length smaller than the eyepiece's (to view tiny nearby objects). Numerical: $\;m_e=1+\dfrac D{f_e}=1+\dfrac{25}5=6\;\Rightarrow\;m_o=\dfrac{30}6=5$ For the objective, with $v_o=-5u_o$ (magnitude-5 real inverted image): $\;\dfrac1{-5u_o}-\dfrac1{u_o}=\dfrac1{1.25}\;\Rightarrow\;\dfrac{-6}{5u_o}=0.8\;\Rightarrow\;u_o=-1.5\text{ cm}$ So the object should be placed 1.5 cm in front of the objective.
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Ray diagram: same general form as Q86, but with the final image formed at the near point rather than at infinity, requiring the eyepiece to be moved slightly closer to the intermediate image than $f_e$ . Radius of the Moon: $\;R_{moon}=\dfrac{3.48\times10^6}2=1.74\times10^6\text{ m}$ ; distance to the Moon, $R_o=3.8\times10^8$ m; $f_o=15$ m. Since the Moon subtends a small angle $\theta$ at the objective, and the image forms essentially at the focus of the objective: $\;\theta=\dfrac{R_{moon}}{R_o}=\dfrac{h}{f_o}$ $\Rightarrow\;h=\dfrac{R_{moon}\times f_o}{R_o}=\dfrac{1.74\times10^6\times15}{3.8\times10^8}\approx6.87\times10^{-2}\text{ m}$ Diameter of the image of the Moon $=2h\approx13.74\times10^{-2}\text{ m}=13.74\text{ cm}$
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Ray diagram: as in Q86 (parallel rays from a distant object focused by the objective, re-collimated by the eyepiece). Angular magnification (normal adjustment): $\;M=\dfrac{f_o}{f_e}$ ; with the final image at the near point: $\;M=\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}D\right)$ To increase $M$ : increase the objective's focal length $f_o$ and decrease the eyepiece's focal length $f_e$ . Achieving large resolution requires a large-aperture objective, since resolving power is proportional to aperture (both for light-gathering and to reduce diffraction-limited blurring). However, large lenses are heavy, difficult to support without sagging under their own weight, and suffer from chromatic and spherical aberration — these limitations are overcome by using a reflecting (mirror) telescope instead of a refracting one.
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(a) Same ray diagram as Q86. (b) Focal lengths: $f_1=\dfrac{100}3\approx33.3$ cm, $f_2=\dfrac{100}6\approx16.7$ cm, $f_3=10$ cm. A telescope objective needs a LARGE focal length and a LARGE aperture (for good light-gathering and resolving power); $L_1$ ( $f=33.3$ cm, aperture 8 cm — both the largest) should be used as the OBJECTIVE. The eyepiece needs a SHORT focal length; $L_3$ ( $f=10$ cm, the shortest) should be used as the EYEPIECE. This choice also maximises the magnifying power $M=f_o/f_e$ .
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Focal lengths: $f_1=\dfrac{100}6\approx16.7$ cm, $f_2=\dfrac{100}3\approx33.3$ cm, $f_3=10$ cm. (i) Telescope: objective needs LARGE aperture and LARGE focal length $\Rightarrow$ use $L_2$ ( $f=33.3$ cm, aperture 8 cm) as the objective; eyepiece needs small focal length $\Rightarrow$ use $L_3$ ( $f=10$ cm) as the eyepiece. (ii) Microscope: both lenses need SHORT focal lengths, with the objective's focal length smaller than the eyepiece's, and small apertures $\Rightarrow$ use $L_3$ ( $f=10$ cm, aperture 1 cm) as the objective and $L_1$ ( $f=16.7$ cm, aperture 1 cm) as the eyepiece.
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(i) Same ray diagram and derivation as Q86/Q89: $\;M=\dfrac{f_o}{f_e}$ (ii) For $L_1$ ( $f_1=+24$ cm): $u_1=-36$ cm. $\;\dfrac1{v_1}=\dfrac1{f_1}+\dfrac1{u_1}=\dfrac1{24}-\dfrac1{36}=\dfrac{3-2}{72}=\dfrac1{72}\;\Rightarrow\;v_1=72\text{ cm}$ So $L_1$ would form its image 72 cm to its right — but $L_2$ is only 45 cm away, so this image forms $72-45=27$ cm beyond $L_2$ , acting as a virtual object for $L_2$ : $\;u_2=+27\text{ cm}$ For $L_2$ ( $f_2=-18$ cm): $\;\dfrac1{v_2}=\dfrac1{f_2}+\dfrac1{u_2}=-\dfrac1{18}+\dfrac1{27}=\dfrac{-3+2}{54}=-\dfrac1{54}\;\Rightarrow\;v_2=-54\text{ cm}$ So the final image forms 54 cm to the left of $L_2$ — since $L_1$ is 45 cm to the left of $L_2$ , this places the final image $54-45=9$ cm to the left of $L_1$ (i.e. on the object's side, between the object and $L_1$ ). This image is virtual. Overall magnification: $\;m=m_1\times m_2=\dfrac{v_1}{u_1}\times\dfrac{v_2}{u_2}=\dfrac{72}{-36}\times\dfrac{-54}{27}=(-2)\times(-2)=4$ Image height: $\;h_i=m\times h_o=4\times1\text{ cm}=4\text{ cm}$ (erect relative to the object, since $m$ is positive; virtual).
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(i) Same ray diagram and derivation as Q86/Q89/Q92: $\;M=\dfrac{f_o}{f_e}$ (ii) $\;M=\dfrac{f_o}{f_e}=2.9\;\Rightarrow\;f_o=2.9f_e$ , and (normal adjustment) $\;L=f_o+f_e=150\text{ cm}$ $\Rightarrow\;2.9f_e+f_e=150\;\Rightarrow\;3.9f_e=150\;\Rightarrow\;f_e\approx38.5\text{ cm}$ $\;f_o=150-38.5=111.5\text{ cm}$
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(a) $\;M=\dfrac{f_o}{f_e}=\dfrac{2000\text{ cm}}{1\text{ cm}}=2000$ (b) $\;R_{moon}=\dfrac{3.5\times10^6}2=1.75\times10^6\text{ m}$ ; $R_o=3.8\times10^8$ m; $f_o=20$ m. $\;h=\dfrac{R_{moon}\times f_o}{R_o}=\dfrac{1.75\times10^6\times20}{3.8\times10^8}\approx9.21\times10^{-2}\text{ m}$ Diameter of the image $=2h\approx18.42\times10^{-2}\text{ m}=18.42\text{ cm}$
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(i) A concave mirror of large aperture gathers much more light and reflects nearly all of it (very little absorption), unlike a lens of similarly large aperture, which absorbs and scatters a significant fraction of the incident light passing through its bulk. Hence the image formed in a reflecting telescope is much brighter, allowing even very distant, faint stars to be observed. (ii) Since mirrors of much larger aperture are practically achievable (a mirror can be supported over its entire back surface), reflecting telescopes have a considerably higher resolving power than refracting telescopes.
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Ray diagram: as in Q86. Magnifying power is the ratio of the angle subtended at the eye by the final image to the angle subtended at the (unaided) eye by the object. (i) Focal lengths: $200$ cm, $25$ cm, $10$ cm respectively. The objective should have the LARGEST focal length (smallest power) $\Rightarrow$ use the $0.5$ D lens ( $f=200$ cm) as the objective. The eyepiece should have the SMALLEST focal length (largest power) $\Rightarrow$ use the $10$ D lens ( $f=10$ cm) as the eyepiece. (This choice also gives the largest possible $M=f_o/f_e$ .) (ii) A large aperture for the objective gives greater light-gathering power (a brighter image of faint, distant stars) and a higher resolving power (ability to distinguish two close objects), since resolving power increases with aperture.
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(a) Same ray diagram as Q86. (b) The objectives of refracting telescopes suffer from chromatic aberration (different colours of light, having different refractive indices, focus at slightly different points, blurring and colour-fringing the image) and spherical aberration (a spherical lens does not bring all parallel rays, at different distances from the axis, to exactly the same focus). In a reflecting telescope, the lens objective is replaced by a paraboloidal concave mirror: since reflection does not depend on colour, chromatic aberration is eliminated entirely; and using a paraboloidal (rather than spherical) mirror shape eliminates spherical aberration as well.
29.↑ back to the question
Ray diagram: as in Q86. Magnifying power, $M=\beta/\alpha$ , is the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ subtended at the (unaided) eye by the distant object. Two factors to increase magnifying power: (i) increase the focal length of the objective, $f_o$ ; (ii) decrease the focal length of the eyepiece, $f_e$ — since $M=f_o/f_e$ . Two main limitations of a refracting telescope: (i) large-aperture lenses are very expensive and difficult to fabricate without defects; (ii) large lenses suffer from chromatic aberration and other distortions, and tend to sag under their own weight (they can only be supported at their edges). These limitations are minimized by using a reflecting-type telescope, where a large concave mirror (which can be supported fully from behind, is free of chromatic aberration, and is cheaper to make defect-free at large sizes) replaces the objective lens.
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(a) Same ray diagram as Q86. For large magnifying power and high resolution: the objective should have a LARGE focal length (for magnification) and a LARGE aperture (for resolution and light-gathering power); the eyepiece should have a SMALL focal length. (b) $u_o=-2.5$ cm, $f_o=1.25$ cm: $\;\dfrac1{v_o}=\dfrac1{f_o}+\dfrac1{u_o}=\dfrac1{1.25}-\dfrac1{2.5}=0.8-0.4=0.4\;\Rightarrow\;v_o=2.5\text{ cm}$ For the final image to form at infinity, the intermediate image (formed by the objective, at $v_o=2.5$ cm) must lie exactly at the focus of the eyepiece: $\;u_e=f_e=5\text{ cm}$ Distance between objective and eyepiece $\;=v_o+u_e=2.5+5=7.5\text{ cm}$