NCERT Exemplar Solutions Class 12 Physics Ray Optics and Optical Instruments — Very Short, Short and Long Answer Questions Q17–32

NCERT Exemplar Problems, Class 12 Physics — Ray Optics and Optical Instruments, Very Short Answer Questions 17 to 21, Short Answer Questions 22 to 27 and Long Answer Questions 28 to 32. All 16 questions solved. The book numbers them 9.17 to 9.32.

Three ideas carry this page. Refraction is Snell’s law applied surface by surface: apparent depth, prisms, the critical angle and the optical-fibre condition all come from $\mu_1\sin\theta_1=\mu_2\sin\theta_2$. Thin-lens and mirror formulas are differentiable: a small change in $u$ gives a small change in $v$, which is how image lengths and conjugate positions are found. Fermat’s principle replaces geometry with calculus: a lens or a gravitational field is just a medium in which the optical path is stationary.

Key insight. Reduce each situation to one angle condition or one stationary-path condition, then solve. In Questions 23, 26 and 28 it is a Snell’s-law equation at a single surface, and in Questions 29, 30 and 32 it is a small deviation obtained from a gradient of optical path.

Very Short Answer Questions

Question 17

Will the focal length of a lens for red light be more, same or less than that for blue light?

Solution. The lens-maker’s formula is $$\frac1f=(\mu-1)\left(\frac1{R_1}-\frac1{R_2}\right).$$ The refractive index of glass is smaller for red light than for blue light, $\mu_{\text{red}}<\mu_{\text{blue}}$, while the radii are the same for both. So $(\mu-1)$ is smaller for red, $1/f$ is smaller, and $f$ is larger.

Question 18

The near vision of an average person is $25$ cm. To view an object with an angular magnification of $10$, what should be the power of the microscope?

Solution. For a simple microscope used with the image at infinity the angular magnification is $m=D/f$, with $D=25$ cm. Then $$f=\frac{D}{m}=\frac{25}{10}=2.5\ \text{cm}=0.025\ \text{m},\qquad P=\frac1f=\frac1{0.025}=40\ \text{D}.$$ If the image is instead formed at the near point, $m=1+D/f$ and the power needed would be $36$ D; the book’s answer uses the first form.

Question 19

An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

Solution. The focal length depends on $\dfrac1{R_1}-\dfrac1{R_2}$. Reversing the lens swaps the two surfaces: the new $R_1$ is the old $-R_2$ and the new $R_2$ is the old $-R_1$, so the difference is $$\frac1{-R_2}-\frac1{-R_1}=\frac1{R_1}-\frac1{R_2},$$ unchanged. Hence $f$ is the same, and for a thin lens the object distance is measured from the same point, so $\dfrac1v-\dfrac1u=\dfrac1f$ gives the same $v$.

Question 20

Three immiscible liquids of densities $d_1>d_2>d_3$ and refractive indices $\mu_1>\mu_2>\mu_3$ are put in a beaker. The height of each liquid column is $\dfrac h3$. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

Solution. Immiscible liquids settle with the densest at the bottom, so the layers from the bottom upwards are $\mu_1,\mu_2,\mu_3$, each of thickness $h/3$. For near-normal viewing, a layer of real thickness $t$ and index $\mu$ seen from air above appears to have thickness $t/\mu$, and the image formed by one layer acts as the object for the next layer above. The apparent depths therefore add: $$\text{apparent depth}=\frac{h/3}{\mu_1}+\frac{h/3}{\mu_2}+\frac{h/3}{\mu_3}.$$

Question 21

For a glass prism ($\mu=\sqrt3$) the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

Solution. At minimum deviation $$\mu=\frac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2}.$$ With $\delta_m=A$ the numerator is $\sin A$, so $$\mu=\frac{\sin A}{\sin\frac A2}=\frac{2\sin\frac A2\cos\frac A2}{\sin\frac A2}=2\cos\frac A2=\sqrt3.$$ Hence $\cos\dfrac A2=\dfrac{\sqrt3}{2}$, so $\dfrac A2=30^\circ$.

Short Answer Questions

Question 22

A short object of length $L$ is placed along the principal axis of a concave mirror away from focus. The object distance is $u$. If the mirror has a focal length $f$, what will be the length of the image? You may take $L\ll|v-f|$.

Solution. Use magnitudes with the object beyond the focus, so the image is real and $\dfrac1u+\dfrac1v=\dfrac1f$. Then $$v=\frac{uf}{u-f}.$$ A short object means $L$ is a small change $\Delta u$ in $u$, which produces a change $\Delta v$ in $v$. Differentiate, or equivalently differentiate the mirror formula directly: $$-\frac1{u^2}\,du-\frac1{v^2}\,dv=0\;\Rightarrow\;|dv|=\frac{v^2}{u^2}\,|du|.$$ With $\dfrac vu=\dfrac f{u-f}$ this is $$L’=\left(\frac{f}{u-f}\right)^2L=\frac{f^2L}{(u-f)^2}.$$ So the longitudinal magnification is the square of the transverse magnification $f/(u-f)$.

The key’s intermediate step prints $(u-f)^2\times L^2/4$ where $(u-f)^2-L^2/4$ is meant; its final result is the same as above.

Question 23

A circular disc of radius $R$ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius $a$ (Fig. 9.5). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index $\mu$ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

Solution. In the figure the disc $AB$ lies at depth $d$ below the plane of the rim, with centre $C$ on the axis below the centre $O$ of the top plane. The eye is at the rim point $M$, at distance $a$ from $O$. Measure every angle from the vertical, which is the normal to the flat top surface.

Empty bowl. The far edge $A$ is at horizontal distance $a+R$ from $M$ and depth $d$ below it. The ray $AM$ travels in air and leaves at angle $\alpha$ to the vertical with $$\sin\alpha=\frac{a+R}{\sqrt{(a+R)^2+d^2}}.$$

Filled bowl. Now the near edge $B$ becomes just visible along the same direction $\alpha$ to the eye. The ray $BM$ lies in the liquid, at horizontal distance $a-R$ from $M$, so its angle $i$ to the vertical satisfies $$\sin i=\frac{a-R}{\sqrt{(a-R)^2+d^2}}.$$ At the liquid surface it emerges at angle $\alpha$, so Snell’s law gives $\mu\sin i=\sin\alpha$: $$\mu\,\frac{a-R}{\sqrt{(a-R)^2+d^2}}=\frac{a+R}{\sqrt{(a+R)^2+d^2}}.$$ Square and clear the denominators: $$\mu^2(a-R)^2\big[(a+R)^2+d^2\big]=(a+R)^2\big[(a-R)^2+d^2\big].$$ Collect the $d^2$ terms: $$d^2\big[\mu^2(a-R)^2-(a+R)^2\big]=(a+R)^2(a-R)^2\,(1-\mu^2),$$ $$d^2=\frac{(a^2-R^2)^2(\mu^2-1)}{(a+R)^2-\mu^2(a-R)^2}.$$

Printed key error. The key takes $\sin\alpha=\dfrac{a+R}{\sqrt{d^2+(a-R)^2}}$, with the wrong horizontal distance inside the root, and ends with $d=\dfrac{\mu(a^2-R^2)}{\sqrt{(a+R)^2-\mu(a-R)^2}}$. A direct check shows the difference: with $a=10$, $R=4$, $d=5$ the geometry gives $\mu=1.2259$; the formula above returns $d=5$, while the key’s formula returns $8.36$.

Question 24

A thin convex lens of focal length $25$ cm is cut into two pieces $0.5$ cm above the principal axis. The top part is placed at $(0,0)$ and an object placed at $(-50\ \text{cm},0)$. Find the coordinates of the image.

Solution. The cut is $0.5$ cm above the original axis, so the top piece has its optical centre $0.5$ cm below its cut edge. That edge sits at the origin, so the optical centre, and the axis of the piece, is at $(0,-0.5)$ cm. The object at $(-50,0)$ is therefore $0.5$ cm above this axis.

Along the axis of the piece, with $u=-50$ cm and $f=+25$ cm, $$\frac1v=\frac1f+\frac1u=\frac1{25}-\frac1{50}=\frac1{50}\;\Rightarrow\;v=50\ \text{cm},\qquad m=\frac vu=-1.$$ The image is $0.5$ cm below the axis of the piece, which is itself $0.5$ cm below the $x$-axis. So the image is at $x=50$ cm and $$y=-0.5-0.5=-1\ \text{cm}.$$

Question 25

In many experimental set-ups the source and screen are fixed at a distance, say $D$, and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

Solution. Let the lens be at distance $u$ from the source, so it is $v=D-u$ from the screen. Writing the lens formula with magnitudes for a real image, $$\frac1u+\frac1{D-u}=\frac1f\;\Rightarrow\;u^2-Du+Df=0.$$ This quadratic in $u$ has two real roots when its discriminant is positive, that is when $D^2-4Df\ge0$, or $D\ge4f$: $$u=\frac{D\pm\sqrt{D^2-4Df}}{2}.$$ So there are two lens positions. Their separation is $$d=\frac{D+\sqrt{D^2-4Df}}2-\frac{D-\sqrt{D^2-4Df}}2=\sqrt{D^2-4Df}.$$ Write $d=\sqrt{D^2-4Df}$. The lens positions are $u_1=\dfrac{D-d}{2}$, $v_1=\dfrac{D+d}2$ and $u_2=\dfrac{D+d}2$, $v_2=\dfrac{D-d}2$, the two being the same pair with object and image interchanged. The magnifications (magnitudes) are $$m_1=\frac{v_1}{u_1}=\frac{D+d}{D-d},\qquad m_2=\frac{v_2}{u_2}=\frac{D-d}{D+d}.$$ Taking the larger image over the smaller (the first position, with the lens nearer the source, gives the larger): $$\frac{\text{larger}}{\text{smaller}}=\frac{D+d}{D-d}\Big/\frac{D-d}{D+d}=\left(\frac{D+d}{D-d}\right)^2.$$

Question 26

A jar of height $h$ is filled with a transparent liquid of refractive index $\mu$ (Fig. 9.6). At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible.

Solution. Light from the dot reaches the top surface at various angles of incidence $i$. A ray that meets the surface at more than the critical angle $c$ is totally reflected and does not leave the liquid. Rays at angles below $c$ emerge and could be seen from above. The disc must therefore cover every point where a ray leaves, which means the points reached by rays at the critical angle $c$ define the edge of the disc. A ray at $c$ meets the surface at horizontal distance $h\tan c$ from the centre, so the disc must have radius at least $h\tan c$.

Here $\sin c=\dfrac1\mu$, so $$\tan c=\frac{\sin c}{\sqrt{1-\sin^2c}}=\frac{1}{\sqrt{\mu^2-1}}.$$ The minimum diameter is $$d=2h\tan c=\frac{2h}{\sqrt{\mu^2-1}}.$$

Question 27

A myopic adult has a far point at $0.1$ m. His power of accommodation is $4$ diopters. (i) What power lenses are required to see distant objects? (ii) What is his near point without glasses? (iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be $2$ cm.)

Solution. (i) To see a distant object, the glasses must form its image at the eye’s far point, $0.1$ m in front of the eye. For a lens the object is at infinity and the image at $v=-0.1$ m, so $$f=-0.1\ \text{m},\qquad P=\frac1f=-10\ \text{D}.$$ A diverging lens of power $-10$ D is needed.

(ii) The eye lens forms images on a retina $2$ cm behind it, a contribution of $1/0.02=50$ D to the power of the focusing system. When relaxed, the eye focuses an object at the far point $0.1$ m: $$P_{\text{relaxed}}=\frac1{0.1}+\frac1{0.02}=10+50=60\ \text{D}.$$ Accommodation adds up to $4$ D, so the maximum power is $64$ D. The nearest object then focused on the retina is at $$\frac1{u}=64-50=14\;\Rightarrow\;u=\frac1{14}\ \text{m}\approx0.07\ \text{m}.$$

(iii) With the $-10$ D glasses (close to the eye) the combined power at full accommodation is $64-10=54$ D, so $$\frac1u=54-50=4\;\Rightarrow\;u=0.25\ \text{m}.$$ With glasses his near point is the normal $25$ cm.

Long Answer Questions

Question 28

Show that for a material with refractive index $\mu\ge\sqrt2$, light incident at any angle shall be guided along a length perpendicular to the incident face.

Solution. Consider a long rod (a light guide) with a flat end face perpendicular to its length. Light enters the end face at angle of incidence $i$ and is refracted at angle $r$: $$\sin i=\mu\sin r.$$ The refracted ray travels towards the side wall and meets it at angle of incidence $90^\circ-r$, since the side wall is perpendicular to the end face. It stays guided, being reflected again and again, only if it is totally internally reflected there: $$\sin(90^\circ-r)=\cos r\ge\frac1\mu.$$ Substitute $\sin r=\dfrac{\sin i}{\mu}$: $$\cos r=\sqrt{1-\frac{\sin^2i}{\mu^2}}\ge\frac1\mu\;\Rightarrow\;\mu^2-\sin^2i\ge1\;\Rightarrow\;\sin^2i\le\mu^2-1.$$ Light must be guided for every angle of incidence up to $i=90^\circ$, where $\sin^2i=1$ is the largest value. The condition then becomes $$1\le\mu^2-1\;\Rightarrow\;\mu\ge\sqrt2.$$ For $\mu\ge\sqrt2$ the inequality holds for all $i$, so every ray entering the end face is totally reflected at the side walls and carried along the length of the guide.

Question 29

The mixture of a pure liquid and a solution in a long vertical column (i.e. horizontal dimensions $\ll$ vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance $d\ll h$, the height of the column.

Solution. Let $y$ be the vertical coordinate, increasing upwards, with refractive index $\mu(y)$. The light speed at height $y$ is $c/\mu(y)$. Consider a narrow horizontal beam of vertical width $\Delta y$, and follow its wavefront, which starts vertical.

After a time $t$ the lower and upper edges of the wavefront have travelled $$s_1=\frac{ct}{\mu(y)},\qquad s_2=\frac{ct}{\mu(y+\Delta y)}\approx\frac{ct}{\mu}-\frac{ct}{\mu^2}\frac{d\mu}{dy}\Delta y.$$ The wavefront therefore tilts through a small angle $$\theta\approx\frac{|s_2-s_1|}{\Delta y}=\frac{ct}{\mu^2}\left|\frac{d\mu}{dy}\right|.$$ The ray, perpendicular to the wavefront, turns by this angle towards the region of higher refractive index. Since $d\ll h$ the gradient and $\mu$ are essentially constant over the path, and the time to travel the horizontal distance $d$ is $t=\dfrac{\mu d}{c}$. Hence $$\theta\approx\frac d\mu\left|\frac{d\mu}{dy}\right|.$$ Taking $\theta$ positive when the ray bends towards lower $y$ (downwards), the sign is carried by the gradient, giving $$\theta\approx-\frac d\mu\,\frac{d\mu}{dy}.$$ In a column that is denser at the bottom, $d\mu/dy<0$ and the ray bends downwards, towards the denser liquid.

Question 30

If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by $$n(r)=1+\frac{2GM}{rc^2},$$ where $r$ is the distance of the point of consideration from the centre of the mass of the massive body, $G$ is the universal gravitational constant, $M$ the mass of the body and $c$ the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.

Solution. Let the ray travel along the line $y=R$, grazing a body of radius $R$ centred at the origin, with $x$ along the ray. At the point $(x,R)$ the distance from the centre is $r=\sqrt{x^2+R^2}$. The index gradient across the ray (in the $y$ direction) is $$\frac{\partial n}{\partial y}=-\frac{2GM}{c^2}\,\frac{y}{r^3}=-\frac{2GM}{c^2}\frac{R}{(x^2+R^2)^{3/2}}.$$ By Question 29 a ray bends towards higher index through $d\theta=\dfrac1n\left|\dfrac{\partial n}{\partial y}\right|dx$ in a path element $dx$. Since $n\approx1$, the total deviation is $$\theta=\int_{-\infty}^{\infty}\frac{2GM}{c^2}\,\frac{R\,dx}{(x^2+R^2)^{3/2}}.$$ With $\displaystyle\int_{-\infty}^{\infty}\frac{dx}{(x^2+R^2)^{3/2}}=\frac2{R^2}$ (substitute $x=R\tan\phi$), $$\theta=\frac{2GMR}{c^2}\cdot\frac2{R^2}=\frac{4GM}{Rc^2}.$$ The ray bends towards the massive body. This is the famous gravitational deflection of light, about $1.75”$ for a ray grazing the Sun.

Question 31

An infinitely long cylinder of radius $R$ is made of an unusual exotic material with refractive index $-1$ (Fig. 9.7). The cylinder is placed between two planes whose normals are along the $y$ direction. The centre of the cylinder $O$ lies along the $y$-axis. A narrow laser beam is directed along the $y$ direction from the lower plate. The laser source is at a horizontal distance $x$ from the diameter in the $y$ direction. Find the range of $x$ such that light emitted from the lower plane does not reach the upper plane.

Solution. The beam travels in the $y$ direction and meets the circular cross-section at a point whose radius makes angle $\theta_i$ with the $y$-axis, where the horizontal offset gives $$\sin\theta_i=\frac xR.$$ The normal at that point is along the radius, so the angle of incidence is $\theta_i$.

Refraction at entry. With $n=-1$, Snell’s law gives $\sin\theta_i=-\sin\theta_r$, so the refracted ray lies on the same side of the normal as the incident ray, at angle $\theta_r=\theta_i$ in magnitude. Compared with a straight path the ray is turned by $2\theta_i$.

Inside. The ray crosses the circle along a chord. By symmetry of the chord, it meets the exit surface at the same angle $\theta_i$ to the normal.

Refraction at exit. The same negative refraction turns it by a further $2\theta_i$ in the same sense. The total deviation of the beam from the $y$ direction is $$\Delta=4\theta_i.$$ The beam leaves the cylinder going upwards, reaching the upper plane, when $\Delta$ lies within $90^\circ$ of straight ahead. It does not reach the upper plane when the deviation turns it to horizontal or downward: $$\frac\pi2<4\theta_i<\frac{3\pi}2\;\Rightarrow\;\frac\pi8<\theta_i<\frac{3\pi}8.$$ (At the limits the emergent ray is exactly parallel to the planes.) Using $x=R\sin\theta_i$, with $\sin\dfrac\pi8=0.383$ and $\sin\dfrac{3\pi}8=0.924$: $$R\sin\frac\pi8<x<R\sin\frac{3\pi}8.$$

Printed key error. The key reaches $\dfrac\pi8\le\sin^{-1}\dfrac xR\le\dfrac{3\pi}8$ correctly, but then drops the sine and prints $\dfrac{R\pi}8\le x\le\dfrac{3\pi R}8$. The upper bound $3\pi R/8=1.18R$ exceeds the radius of the cylinder, so it cannot be right; $x$ must be taken as $R\sin\theta_i$.

Question 32

(i) Consider a thin lens placed between a source (S) and an observer (O) (Fig. 9.8). Let the thickness of the lens vary as $w(b)=w_0-\dfrac{b^2}{\alpha}$, where $b$ is the vertical distance from the pole and $w_0$ is a constant. Using Fermat’s principle, i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.

(ii) A gravitational lens may be assumed to have a varying width of the form $$w(b)=k_1\ln\left(\frac{k_2}{b}\right)\quad(b_{\min}<b<b_{\max}),\qquad w(b)=k_1\ln\left(\frac{k_2}{b_{\min}}\right)\quad(b<b_{\min}).$$ Show that an observer will see an image of a point object as a ring about the centre of the lens with an angular radius $$\beta=\sqrt{\frac{(n-1)k_1\dfrac uv}{u+v}}.$$

Solution. (i) The source $S$ is at distance $u$ from the lens and the observer $O$ at distance $v$ on the other side. A ray from $S$ crosses the lens at height $b$ above the pole $P$. The time of transit is proportional to the optical path, which is the geometric path in air plus the extra optical path $(n-1)w(b)$ added by the glass of thickness $w(b)$: $$\text{OPL}(b)=\sqrt{u^2+b^2}+\sqrt{v^2+b^2}+(n-1)\,w(b).$$ For paraxial rays $b\ll u,v$, so $\sqrt{u^2+b^2}\approx u+\dfrac{b^2}{2u}$ and likewise for $v$: $$\text{OPL}\approx u+v+\frac{b^2}2\left(\frac1u+\frac1v\right)+(n-1)\left(w_0-\frac{b^2}\alpha\right).$$ All paraxial rays arrive at $O$ in the same time, as they must if they converge there, only when the $b^2$ terms cancel: $$\frac12\left(\frac1u+\frac1v\right)-\frac{n-1}\alpha=0\;\Rightarrow\;\frac1u+\frac1v=\frac{2(n-1)}\alpha.$$ Writing $\dfrac1D=\dfrac1u+\dfrac1v$, so that $D=\dfrac{uv}{u+v}$, the condition is $\alpha=2(n-1)D$. Comparing with the lens formula $\dfrac1u+\dfrac1v=\dfrac1f$, the focal length is $$f=\frac{\alpha}{2(n-1)}.$$

(ii) Now the width is $w=k_1\ln\dfrac{k_2}b$ and the optical path is $$\text{OPL}(b)\approx u+v+\frac{b^2}2\left(\frac1u+\frac1v\right)+(n-1)k_1\ln\frac{k_2}b.$$ Fermat’s principle says the path is stationary: $$\frac{d(\text{OPL})}{db}=b\left(\frac1u+\frac1v\right)-\frac{(n-1)k_1}b=0\;\Rightarrow\;b^2=\frac{(n-1)k_1}{\dfrac1u+\dfrac1v}=\frac{(n-1)k_1\,uv}{u+v}.$$ So rays are received only from the circle of radius $b$ about the centre of the lens, and by symmetry from every point of that circle. The observer therefore sees a ring. Its angular radius, seen at distance $v$, is $$\beta=\frac bv=\sqrt{\frac{(n-1)k_1\,uv}{(u+v)\,v^2}}=\sqrt{\frac{(n-1)k_1\,\dfrac uv}{u+v}}.$$

The key writes $w_0+b^2/\alpha$ for $w_0-b^2/\alpha$ in its total-time expression; the result is unaffected.