Ray Optics and Optical Instruments

1. 1Reflection of Light by Spherical Mirrors10 board questions
2. 2Refraction1 board question
3. 3Total Internal Reflection3 board questions
4. 4Refraction at Spherical Surfaces and by Lenses36 board questions
5. 5Refraction through a Prism19 board questions
6. 6Optical Instruments30 board questions

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

1. 1

   Derivation 1

   Use the mirror equation to show that an object placed between $f$ and $2f$ of a concave mirror produces a real image beyond $2f$.

   Asked in: CBSE Class 12 Physics — SA · Delhi 2015 · Section B, 2 marks

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   For a concave mirror, $f<0$. Let the object lie between $f$ and $2f$, i.e. $2f<u<f$ (both negative), which can be written as $\dfrac{1}{2f}>\dfrac1u>\dfrac1f\;$ or $\;-\dfrac{1}{2f}<-\dfrac1u<-\dfrac1f$. From the mirror formula, $\dfrac1v=\dfrac1f-\dfrac1u$. Since $2f<u<f$: subtracting $\dfrac1u$ (using the above inequality) from $\dfrac1f$ gives $\;\dfrac1f-\dfrac1{2f}<\dfrac1f-\dfrac1u<\dfrac1f-\dfrac1f\;$ i.e. $\;\dfrac{1}{2f}<\dfrac1v<0$ This implies $v<0$ (real image) so that $2f>v$ i.e. $|v|>|2f|$, meaning the image is formed beyond $2f$ (and the inequality $\dfrac1v<0$ together with $\dfrac1{2f}<\dfrac1v$ forces $|v|>2|f|$ since both are negative). Hence, an object placed between $f$ and $2f$ of a concave mirror produces a real image formed beyond $2f$.

2. 2

   Derivation 2

   An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the image formation and hence derive the mirror equation $\dfrac1f=\dfrac1u+\dfrac1v$.

   Asked in: CBSE Class 12 Physics — Long Answer · 3/5, 2020 · Section E, 5 marks

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   Consider an object $AB$ placed on the principal axis of a concave mirror between its pole $P$ and focus $F$. A ray $AM$ from the object, parallel to the principal axis, strikes the mirror at $M$ and, after reflection, appears to come from the focus $F$ (i.e. the reflected ray, extended backwards, passes through $F$). A second ray $AP$ incident at the pole $P$ reflects such that $\angle APB=\angle B'PA'$ (laws of reflection). The two reflected rays, when produced backwards, meet at a point $A'$ behind the mirror — this is the virtual, erect, magnified image of $A$; likewise, $A'B'$ is the virtual image of $AB$. Using the Cartesian sign convention: object distance $BP=-u$, image distance $PB'=v$ (note: since the image is virtual, $B'$ lies on the same side as the object's ray extension, so with the convention used, $PB'=v$ is taken as measured to the virtual image point), focal length $FP=-f$, and radius of curvature $CP=-R=-2f$. From similar triangles $\triangle ABC\sim\triangle A'B'C$: $\;\dfrac{AB}{A'B'}=\dfrac{CB}{CB'}=\dfrac{CP-BP}{CP+PB'}=\dfrac{-2f+u}{-2f+v}$ From similar triangles $\triangle MPF\sim\triangle A'B'F$: $\;\dfrac{MP}{A'B'}=\dfrac{FP}{FB'}=\dfrac{FP}{FP+PB'}\;$ i.e. $\;\dfrac{AB}{A'B'}=\dfrac{-f}{-f+v}$ (since $MP=AB$, both being perpendicular heights at the pole and at the object). Equating the two expressions for $AB/A'B'$: $\;\dfrac{-2f+u}{-2f+v}=\dfrac{-f}{-f+v}$ Cross-multiplying and simplifying (expanding both sides and cancelling the $2f^2$ terms) leads to $\;uv=fv+fu\;$, i.e. dividing throughout by $uvf$: $\;\dfrac1f=\dfrac1u+\dfrac1v$ This proves the mirror equation for a concave mirror forming a virtual image.

3. 3

   Derivation 3

   Using lens maker's formula, derive the thin lens formula $\dfrac1f=\dfrac1v-\dfrac1u$.

   Asked in: CBSE Class 12 Physics — SA · 2020 · Section B, 2 marks

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   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.

4. 4

   Derivation 4

   (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.

   Asked in: CBSE Class 12 Physics — Long Answer · 2020 · Section E, 5 marks

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   (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.

5. 5

   Derivation 5

   (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?

   Asked in: CBSE Class 12 Physics — Long Answer · AI 2016 · Section E, 5 marks

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   (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.

6. 6

   Derivation 6

   (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.

   Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2015, Ev · Section E, 5 marks

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   (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).

7. 7

   Derivation 7

   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.

   Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2014C · Section E, 5 marks

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   [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 $\dfrac1f=(n_{21}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$.

8. 8

   Derivation 8

   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 $y$.

   Asked in: CBSE Class 12 Physics — SA · 2018, Cr · Section C, 3 marks

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   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 $\mu_l$): $\;\dfrac1{f_2}=(\mu_l-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=(\mu_l-1)\left(-\dfrac1R\right)$ $\Rightarrow\dfrac{y-x}{xy}=-\dfrac{(\mu_l-1)}{R}\;\Rightarrow\;\mu_l-1=\dfrac{(x-y)R}{xy}$ $$\mu_l=1+\dfrac{(x-y)R}{xy}$$

9. 9

   Derivation 9

   (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.

   Asked in: CBSE Class 12 Physics — Long Answer · AI 2019, Ev · Section E, 5 marks

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   (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).

10. 10

    Derivation 10

    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.

    Asked in: CBSE Class 12 Physics — Long Answer · 2/5, AI 2015 · Section E, 5 marks

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    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 $P=1/f$: $\;P_T=P_1+P_2$

11. 11

    Derivation 11

    Figure shows a ray of light PQ passing through a prism ABC (apex angle $A$), refracting at Q on face AB and again at R on face AC, with the refracted ray QR parallel to the base BC. If the refracted ray QR is parallel to the base BC, show that (i) $r_1=r_2=A/2$, (ii) the angle of minimum deviation is $\delta_m=2i-A$.

    Asked in: CBSE Class 12 Physics — SA · Foreign 2014 · Section C, 3 marks

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    (i) Since QR is parallel to the base BC, by geometry the two base angles of the triangle formed by QR and the normals are equal, which forces the ray to be travelling symmetrically inside the prism: the angle of incidence at the first face equals the angle of emergence at the second face, i.e. $i=e$, and correspondingly $r_1=r_2$. Since $r_1+r_2=A$ (the prism relation) and $r_1=r_2$: $\;2r_1=A\;\Rightarrow\;r_1=r_2=\dfrac A2$ (ii) The general relation for a prism is $A+\delta=i+e$, where $\delta$ is the angle of deviation. When the ray travels symmetrically ($i=e$), the deviation is minimum, i.e. $\delta=\delta_m$: $\;A+\delta_m=i+i=2i$ $\Rightarrow\;\delta_m=2i-A$

12. 12

    Derivation 12

    (a) Plot a graph showing the variation of the angle of deviation as a function of the angle of incidence, for light passing through a triangular prism. (b) Derive the relation for the refractive index of the material of the prism in terms of the angle of minimum deviation and the angle of the prism.

    Asked in: CBSE Class 12 Physics — Long Answer · AI 2019 · Section E, 5 marks

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    (a) As the angle of incidence $i$ is increased from a small value, the angle of deviation $\delta$ first decreases, reaches a single minimum value $\delta_m$ (at some particular $i$, where the ray passes symmetrically through the prism, $i=e$), and then increases again. The $\delta$-versus-$i$ graph is thus a smooth curve, concave-up, with a single minimum — it is NOT symmetric about the minimum in general shape but does have $i=e$ at that minimum. (b) For a prism of apex angle $A$: $\;A=r_1+r_2\qquad\text{...(i)}$, and the total deviation $\;\delta=(i-r_1)+(e-r_2)=(i+e)-(r_1+r_2)=(i+e)-A\qquad\text{...(ii)}$ At minimum deviation, $\delta=\delta_m$, $i=e$, and correspondingly $r_1=r_2=r$ (by symmetry). From (i): $\;r=\dfrac A2$ From (ii): $\;\delta_m=2i-A\;\Rightarrow\;i=\dfrac{A+\delta_m}2$ The refractive index of the prism material is then: $\;\mu=\dfrac{\sin i}{\sin r}=\dfrac{\sin\left(\dfrac{A+\delta_m}2\right)}{\sin\left(\dfrac A2\right)}$

13. 13

    Derivation 13

    (a) A ray PQ of light is incident on face AB of a glass prism ABC and emerges out of face AC. Trace the path of the ray and show that $\angle i+\angle e=\angle A+\angle\delta$, where $\delta$ and $e$ denote the angle of deviation and the angle of emergence respectively. Plot a graph showing the variation of the angle of deviation as a function of the angle of incidence. State the condition under which $\angle\delta$ is minimum. (b) Find the relation between the refractive index ($\mu$) of the glass prism and $\angle A$, for the case when the angle of the prism $A$ is equal to the angle of minimum deviation $\delta_m$. Hence obtain the value of the refractive index for a prism angle $A=60°$.

    Asked in: CBSE Class 12 Physics — Long Answer · AI 2015, Ev · Section E, 5 marks

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    (a) At face AB, the deviation produced is $\delta_1=i-r_1$. At face AC, the deviation produced is $\delta_2=e-r_2$. The total deviation is $\;\delta=\delta_1+\delta_2=(i-r_1)+(e-r_2)=(i+e)-(r_1+r_2)$ Since $r_1+r_2=A$: $\;\delta=(i+e)-A\;\Rightarrow\;i+e=A+\delta$ Graph: same $\delta$-versus-$i$ curve as in Q61/Q68 — decreasing then increasing, with a single minimum. The deviation $\delta$ is minimum when the ray traverses the prism symmetrically, i.e. when $i=e$ (equivalently $r_1=r_2$). (b) When $A=\delta_m$: from $\delta_m=2i-A$, we get $\;A=2i-A\;\Rightarrow\;i=A$, and (since $r_1=r_2=A/2$ at minimum deviation) $\;r=\dfrac A2$ $\;\mu=\dfrac{\sin i}{\sin r}=\dfrac{\sin A}{\sin(A/2)}$ For $A=60°$: $\;\mu=\dfrac{\sin60°}{\sin30°}=\dfrac{\sqrt3/2}{1/2}=\sqrt3\approx1.732$

14. 14

    Derivation 14

    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.

    Asked in: CBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marks

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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 $\;m\approx\dfrac L{f_o}\left(1+\dfrac D{f_e}\right)$

15. 15

    Derivation 15

    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).

    Asked in: CBSE Class 12 Physics — Long Answer · AI 2015C · Section E, 5 marks

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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: $\;M=m_o\times m_e=\dfrac L{f_o}\times\dfrac D{f_e}$

16. 16

    Derivation 16

    (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.

    Asked in: CBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marks

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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)

17. 17

    Derivation 17

    (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.

    Asked in: CBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marks

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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).

18. 18

    Derivation 18

    (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.

    Asked in: CBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marks

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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) $\;L=f_o+f_e=150\text{ cm}$ $\Rightarrow\;2.9f_e+f_e=150\;\Rightarrow\;3.9f_e=150\;\Rightarrow\;f_e\approx38.5\text{ cm}$ $\;f_o=150-38.5=111.5\text{ cm}$