Exercises — Ray Optics and Optical Instruments

1. 13 marksNCERT Exercises, Chapter 9

   A small candle, 2.5 cm in size, is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

   Hint. Convert the radius of curvature to a focal length first, then apply the mirror formula with the Cartesian sign convention.

   For a concave mirror the focal length is half the radius of curvature, and both are negative in the Cartesian convention:

   $f=-\frac{R}{2}=-\frac{36}{2}=-18\text{cm},u=-27\text{cm}$

   Applying the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$:

   $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{18}+\frac{1}{27}=\frac{-3+2}{54}=-\frac{1}{54}$

   $v=-54\text{cm}$

   The negative sign means the image forms 54 cm in front of the mirror, on the same side as the object, so it is real and can be caught on a screen placed there.

   The magnification is:

   $m=-\frac{v}{u}=-\frac{-54}{-27}=-2$

   The negative value means the image is inverted, and its size is $2.5\times 2=5.0$ cm, so it is enlarged.

   Moving the candle closer to the mirror decreases $|u|$ towards the focal length. Since $v$ grows without limit as $u$ approaches $f=-18$ cm, the screen must be moved farther away from the mirror. Once the candle is nearer than 18 cm the image becomes virtual and no screen position works at all.

   ✦ v = -54 cm; the image is real, inverted and 5.0 cm tall; moving the candle closer pushes the screen farther away

2. 23 marksNCERT Exercises, Chapter 9

   A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

   Hint. A convex mirror has a positive focal length in the Cartesian convention; check the sign of v to classify the image.

   For a convex mirror the focal length is positive, and the object distance is negative:

   $f=+15\text{cm},u=-12\text{cm}$

   From the mirror formula:

   $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}$

   $v=\frac{60}{9}=6.7\text{cm}$

   The positive value places the image behind the mirror, so it is virtual.

   The magnification is:

   $m=-\frac{v}{u}=-\frac{6.67}{-12}=+0.56$

   Being positive and less than one, the image is erect and diminished, of size $4.5\times 0.56=2.5$ cm.

   As the needle is moved farther away, $|u|$ increases. In the limit $u\to -\infty$ the formula gives $v\to f=+15$ cm and $m\to 0$. So the image moves steadily back from the pole towards the focus while shrinking towards a point. A convex mirror therefore always keeps its virtual image between the pole and the focus, which is why it gives such a wide field of view.

   ✦ v = +6.7 cm behind the mirror; virtual, erect, diminished, m = +0.56, size 2.5 cm; as the needle recedes the image moves towards the focus and shrinks

3. 33 marksNCERT Exercises, Chapter 9

   A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

   Hint. Refractive index is the ratio of real depth to apparent depth; compute the new apparent depth and take the difference.

   The refractive index relates real and apparent depth by:

   $n=\frac{\text{real depth}}{\text{apparent depth}}=\frac{12.5}{9.4}=1.33$

   With the liquid of refractive index 1.63 filled to the same height, the new apparent depth is:

   $d^{\prime }=\frac{12.5}{1.63}=7.67\text{cm}$

   The needle now appears shallower than before, because the denser liquid bends the emerging rays more strongly. The microscope must therefore be lowered by:

   $\Delta d=9.4-7.67=1.73\text{cm}$

   Since the image has risen towards the surface, the microscope is moved downward by about 1.7 cm to refocus on it.

   ✦ n = 1.33; the microscope must be moved down by about 1.73 cm

4. 43 marksNCERT Exercises, Chapter 9

   Figures 9.27(a) and (b) show refraction of a ray in air incident at $60^{\circ }$ with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is $45^{\circ }$ with the normal to a water-glass interface [Fig. 9.27(c)].

   Hint. Get the absolute refractive index of each medium from the first two figures, then form the relative index for the water-to-glass interface.

   From Fig. 9.27(a), light in air at $60^{\circ }$ refracts into glass at $35^{\circ }$:

   $n_g=\frac{\sin 60^{\circ }}{\sin 35^{\circ }}=\frac{0.8660}{0.5736}=1.51$

   From Fig. 9.27(b), light in air at $60^{\circ }$ refracts into water at $47^{\circ }$:

   $n_w=\frac{\sin 60^{\circ }}{\sin 47^{\circ }}=\frac{0.8660}{0.7314}=1.18$

   For the water-to-glass interface of Fig. 9.27(c), Snell's law in the form $n_w\sin i=n_g\sin r$ gives:

   $\sin r=\frac{n_w}{n_g}\sin 45^{\circ }=\frac{1.18}{1.51}\times 0.7071=0.5546$

   $r={\sin }^{-1}(0.5546)=33.7^{\circ }$

   Note that the value $n_w=1.18$ implied by Fig. 9.27(b) is not the real refractive index of water, which is 1.33 and is the figure used in Exercises 9.3 and 9.5 of this same chapter. The angles printed in Fig. 9.27 are illustrative, so this question must be answered from the data it supplies rather than from the physical value.

   ✦ r = 33.7 degrees, using the relative index n_g/n_w implied by the figure

5. 53 marksNCERT Exercises, Chapter 9

   A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

   Hint. Only rays striking the surface within the critical angle escape, so the illuminated patch is a circle whose radius follows from the critical angle.

   Light escapes only where it meets the surface at less than the critical angle. Beyond that it is totally internally reflected back into the water.

   The critical angle satisfies:

   $\sin C=\frac{1}{n}=\frac{1}{1.33}=0.7519⟹C=48.75^{\circ }$

   Rays leaving the bulb within a cone of half-angle $C$ reach the surface inside a circle of radius:

   $R=h\tan C=0.80\times \tan 48.75^{\circ }=0.80\times 1.140=0.91\text{m}$

   The area of this circular patch is:

   $A=\pi R^2=\pi (0.91{)}^2=2.6{\text{m}}^2$

   Outside this circle the surface looks like a mirror from below, because every ray there strikes it beyond the critical angle.

   ✦ Area = 2.6 m^2, a circle of radius about 0.91 m

6. 63 marksNCERT Exercises, Chapter 9

   A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be $40^{\circ }$. What is the refractive index of the material of the prism? The refracting angle of the prism is $60^{\circ }$. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

   Hint. Use the prism formula for minimum deviation, then repeat it with the relative refractive index of glass with respect to water.

   At minimum deviation the prism formula is:

   $n=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

   Substituting $A=60^{\circ }$ and $D_m=40^{\circ }$:

   $n=\frac{\sin 50^{\circ }}{\sin 30^{\circ }}=\frac{0.7660}{0.5}=1.53$

   In water the prism no longer bends light as strongly, because what matters is the relative refractive index:

   $n_{gw}=\frac{n_g}{n_w}=\frac{1.53}{1.33}=1.15$

   Applying the same formula in reverse to find the new minimum deviation:

   $\sin \left(\frac{60^{\circ }+D_m^{\prime }}{2}\right)=n_{gw}\sin 30^{\circ }=1.15\times 0.5=0.5754$

   $\frac{60^{\circ }+D_m^{\prime }}{2}=35.2^{\circ }⟹D_m^{\prime }=10.3^{\circ }$

   The deviation drops sharply, from $40^{\circ }$ to about $10^{\circ }$, since the contrast in optical density between glass and water is far smaller than between glass and air.

   ✦ n = 1.53; the new angle of minimum deviation in water is about 10.3 degrees

7. 72 marksNCERT Exercises, Chapter 9

   Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

   Hint. Apply the lens maker's formula, noting that for a double convex lens with equal curvatures the two radii are equal in magnitude but opposite in sign.

   The lens maker's formula is:

   $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

   For a double convex lens with both faces of the same radius, $R_1=+R$ and $R_2=-R$, so:

   $\frac{1}{f}=(n-1)\left(\frac{1}{R}+\frac{1}{R}\right)=\frac{2(n-1)}{R}$

   Rearranging for $R$ and substituting $n=1.55$, $f=20$ cm:

   $R=2f(n-1)=2\times 20\times 0.55=22\text{cm}$

   The opposite signs of $R_1$ and $R_2$ are what make the two surfaces reinforce rather than cancel, so forgetting them halves the answer.

   ✦ R = 22 cm

8. 84 marksNCERT Exercises, Chapter 9

   A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?

   Hint. The beam is already converging towards P, so P acts as a virtual object with a positive object distance.

   Because the light is converging towards P before it reaches the lens, P is a virtual object. In the Cartesian convention this makes the object distance positive:

   $u=+12\text{cm}$

   (a) For the convex lens, $f=+20$ cm. The lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ gives:

   $\frac{1}{v}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}$

   $v=7.5\text{cm}$

   The converging lens pulls the convergence point nearer, from 12 cm to 7.5 cm beyond the lens.

   (b) For the concave lens, $f=-16$ cm:

   $\frac{1}{v}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48}$

   $v=48\text{cm}$

   The diverging lens pushes the convergence point away, from 12 cm out to 48 cm. In both cases $v$ is positive, so the beam still converges to a real point on the far side.

   ✦ (a) 7.5 cm from the lens (b) 48 cm from the lens, both on the far side

9. 93 marksNCERT Exercises, Chapter 9

   An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

   Hint. A concave lens has a negative focal length; check the sign of v and the size of m to classify the image.

   For a concave lens the focal length is negative:

   $f=-21\text{cm},u=-14\text{cm}$

   From the lens formula:

   $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{21}-\frac{1}{14}=\frac{-2-3}{42}=-\frac{5}{42}$

   $v=-8.4\text{cm}$

   The negative sign puts the image on the same side as the object, so it is virtual.

   The magnification is:

   $m=\frac{v}{u}=\frac{-8.4}{-14}=+0.6$

   Being positive and less than one, the image is erect and diminished, of size $3.0\times 0.6=1.8$ cm.

   As the object is moved further away, $|v|$ increases towards the focal length while $m$ falls towards zero. So the image moves away from the lens towards the focus at 21 cm and becomes steadily smaller. A concave lens always produces a virtual, erect, diminished image, wherever the object is placed.

   ✦ v = -8.4 cm; virtual, erect, diminished, size 1.8 cm; as the object recedes the image moves towards the focus and shrinks

10. 102 marksNCERT Exercises, Chapter 9

    What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

    Hint. For thin lenses in contact the reciprocals of the focal lengths add; the sign of the result classifies the combination.

    For two thin lenses in contact the combined focal length satisfies:

    $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$

    Taking $f_1=+30$ cm for the convex lens and $f_2=-20$ cm for the concave lens:

    $\frac{1}{F}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}$

    $F=-60\text{cm}$

    The focal length is negative, so the combination behaves as a diverging lens. This happens because the concave lens is the stronger of the two: its power is $-5$ D against the convex lens's $+3.33$ D, giving a net power of $-1.67$ D. Adding powers rather than focal lengths is the faster route and makes the sign obvious.

    ✦ F = -60 cm; the combination is diverging

11. 115 marksNCERT Exercises, Chapter 9

    A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

    Hint. Work backwards from the eyepiece: find where its object must sit, subtract from the tube length to get the objective's image distance, then use the lens formula.

    Work backwards from the eyepiece in each case.

    (a) Final image at 25 cm. For the eyepiece, $v_e=-25$ cm and $f_e=6.25$ cm:

    $\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{25}-\frac{1}{6.25}=-\frac{1+4}{25}=-\frac{1}{5}$

    $u_e=-5\text{cm}$

    The objective's image must therefore lie 5 cm before the eyepiece, so its distance from the objective is $v_o=15-5=10$ cm. From the lens formula for the objective:

    $\frac{1}{u_o}=\frac{1}{v_o}-\frac{1}{f_o}=\frac{1}{10}-\frac{1}{2}=-\frac{4}{10}$

    $u_o=-2.5\text{cm}$

    The magnifying power is:

    $M=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}\right)=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=4\times 5=20$

    (b) Final image at infinity. The eyepiece's object must then sit exactly at its focus, so $|u_e|=f_e=6.25$ cm and $v_o=15-6.25=8.75$ cm:

    $\frac{1}{u_o}=\frac{1}{8.75}-\frac{1}{2}=-0.3857⟹u_o=-2.59\text{cm}$

    $M=\frac{v_o}{|u_o|}\cdot \frac{D}{f_e}=\frac{8.75}{2.59}\times \frac{25}{6.25}=3.38\times 4=13.5$

    The magnification is lower with the image at infinity, but the eye is relaxed, which is why that setting is preferred for long viewing.

    ✦ (a) object at 2.5 cm, M = 20 (b) object at 2.59 cm, M = 13.5

12. 124 marksNCERT Exercises, Chapter 9

    A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

    Hint. Find the objective's image distance first, then the eyepiece's object distance for a final image at the near point, and add them.

    Work in centimetres: $f_o=0.8$ cm, $u_o=-0.9$ cm, $f_e=2.5$ cm, $D=25$ cm.

    For the objective:

    $\frac{1}{v_o}=\frac{1}{f_o}+\frac{1}{u_o}=\frac{1}{0.8}-\frac{1}{0.9}=1.25-1.111=0.1389$

    $v_o=7.2\text{cm}$

    For the eyepiece, the final image is at the near point, so $v_e=-25$ cm:

    $\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{2.5}=-0.44⟹u_e=-2.27\text{cm}$

    The separation between the lenses is the sum of these two distances:

    $L=v_o+|u_e|=7.2+2.27=9.47\text{cm}$

    The separation comes out larger than the objective's image distance alone because the eyepiece must sit far enough back that its own object falls just inside its focal length.

    The magnifying power is the product of the objective's linear magnification and the eyepiece's angular magnification, since the two lenses act in series:

    $M=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}\right)=\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)=8\times 11=88$

    ✦ Separation = 9.47 cm; magnifying power = 88

13. 132 marksNCERT Exercises, Chapter 9

    A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

    Hint. In normal adjustment the magnifying power is the ratio of the focal lengths and the tube length is their sum.

    In normal adjustment, with the final image at infinity, the magnifying power of a telescope is:

    $M=\frac{f_o}{f_e}=\frac{144}{6.0}=24$

    The separation between the lenses is the sum of the focal lengths, because the objective's image forms at the common focal point of the two lenses:

    $L=f_o+f_e=144+6.0=150\text{cm}$

    Note how the two requirements pull in opposite directions: a long objective focal length gives high magnification but also makes the instrument long, which is why large refracting telescopes are physically enormous.

    ✦ M = 24; separation = 150 cm

14. 144 marksNCERT Exercises, Chapter 9

    (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48\times 10^6$ m, and the radius of lunar orbit is $3.8\times 10^8$ m.

    Hint. Use the focal length ratio for the magnification; for the image size, find the angle the moon subtends and multiply by the objective's focal length.

    (a) With both focal lengths in the same unit, $f_o=15$ m and $f_e=1.0$ cm $=0.01$ m:

    $M=\frac{f_o}{f_e}=\frac{15}{0.01}=1500$

    (b) The moon subtends an angle at the objective given by its diameter divided by its distance:

    $\theta =\frac{3.48\times 10^6}{3.8\times 10^8}=9.16\times 10^{-3}\text{rad}$

    The objective forms the image at its focal plane, so the image diameter is:

    $d=f_o\theta =15\times 9.16\times 10^{-3}=0.137\text{m}=13.7\text{cm}$

    The answer depends only on the angle the moon subtends, not on its actual size or distance separately, which is why a small angle times a long focal length still gives a usefully large image.

    ✦ (a) M = 1500 (b) image diameter = 0.137 m, about 13.7 cm

15. 155 marksNCERT Exercises, Chapter 9

    Use the mirror equation to deduce that: (a) an object placed between $f$ and $2f$ of a concave mirror produces a real image beyond $2f$; (b) a convex mirror always produces a virtual image independent of the location of the object; (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole; (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

    Hint. Rearrange the mirror equation to isolate v, then reason about the sign and size of each expression over the stated range of u.

    Write the mirror equation as $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ throughout.

    (a) For a concave mirror $f<0$. Let the object lie between $f$ and $2f$, so $f>u>2f$ in signed terms, that is $|f|<|u|<2|f|$. Then $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ is negative, so $v$ is negative and the image is real. Comparing magnitudes, as $|u|$ runs from $|f|$ to $2|f|$, $|v|$ runs from infinity down to $2|f|$. Hence $|v|>2|f|$ throughout, so the image lies beyond $2f$.

    (b) For a convex mirror $f>0$, and a real object always has $u<0$, so $-\frac{1}{u}>0$. Both terms on the right are then positive, giving $\frac{1}{v}>0$ and therefore $v>0$ for every object position. A positive $v$ places the image behind the mirror, so it is always virtual.

    (c) From $\frac{1}{v}=\frac{1}{f}+\frac{1}{|u|}$, the right-hand side always exceeds $\frac{1}{f}$, so $v<f$: the image lies between the pole and the focus. The magnification is $m=-\frac{v}{u}=\frac{v}{|u|}$, and since $v<f<|u|$ is not guaranteed directly, note instead that $v=\frac{f|u|}{f+|u|}$, so $m=\frac{f}{f+|u|}<1$ always. The image is therefore always diminished.

    (d) For a concave mirror with the object between pole and focus, $|u|<|f|$. Then $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{|f|}+\frac{1}{|u|}$, which is positive because $|u|<|f|$. So $v>0$ and the image is virtual. Its magnification $m=\frac{v}{|u|}$ works out as $\frac{|f|}{|f|-|u|}>1$, so the image is enlarged.

    Each conclusion follows from the sign and magnitude of a single algebraic expression, which is why the exercise is set as an alternative to drawing ray diagrams.

    ✦ All four results follow from the sign and magnitude of 1/v = 1/f - 1/u over the stated range of u

16. 163 marksNCERT Exercises, Chapter 9

    A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

    Hint. Use the normal shift produced by a parallel-sided slab, and check whether the slab's position enters the expression.

    A parallel-sided slab of thickness $t$ and refractive index $n$ raises the apparent position of an object by the normal shift:

    $\text{shift}=t\left(1-\frac{1}{n}\right)$

    Substituting $t=15$ cm and $n=1.5$:

    $\text{shift}=15\left(1-\frac{1}{1.5}\right)=15\times \frac{1}{3}=5\text{cm}$

    The pin appears raised by 5 cm.

    The answer does not depend on where the slab is placed. The expression contains only the thickness and the refractive index, with no term for the slab's distance from either the pin or the observer. Physically, the emergent ray from a parallel-sided slab is displaced sideways but travels in its original direction, and that lateral displacement is fixed by the slab alone. Moving the slab up or down therefore shifts where the ray crosses the gap but not by how much it is offset, which is why the apparent rise stays at 5 cm. Note that the 50 cm viewing distance is not needed either.

    ✦ The pin appears raised by 5 cm; the answer is independent of the slab's position

17. 174 marksNCERT Exercises, Chapter 9

    (a) Figure 9.28 shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place? (b) What is the answer if there is no outer covering of the pipe?

    Hint. Find the critical angle at the core-cladding boundary, convert it to the angle at the entry face, then repeat with air as the outer medium.

    (a) At the core-cladding boundary the critical angle satisfies:

    $\sin C=\frac{n_{\text{cladding}}}{n_{\text{core}}}=\frac{1.44}{1.68}=0.8571⟹C=59^{\circ }$

    For total internal reflection the ray must strike that wall at an angle greater than $C$. A ray refracted into the fibre at angle $r$ to the axis meets the wall at $90^{\circ }-r$, so the condition is $90^{\circ }-r>C$, that is $r<31^{\circ }$.

    The largest angle of incidence at the entry face follows from Snell's law:

    $\sin i_{\max }=n_{\text{core}}\sin 31^{\circ }=1.68\times 0.515=0.865⟹i_{\max }\approx 60^{\circ }$

    So all rays entering within about $60^{\circ }$ of the axis are totally reflected, and the usable range is $0^{\circ }$ to $60^{\circ }$.

    (b) Without the covering the outer medium is air, so:

    $\sin C^{\prime }=\frac{1}{1.68}=0.5952⟹C^{\prime }=36.5^{\circ }$

    The condition becomes $r<90^{\circ }-36.5^{\circ }=53.5^{\circ }$. But the largest possible refraction angle inside the fibre is ${\sin }^{-1}(1/1.68)=36.5^{\circ }$, which is already less than $53.5^{\circ }$. The condition is therefore satisfied automatically, so every ray entering the pipe at any angle undergoes total internal reflection.

    ✦ (a) rays within about 60 degrees of the axis (b) all rays entering the pipe, at any angle

18. 183 marksNCERT Exercises, Chapter 9

    The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

    Hint. For a real image on a screen, the object-to-image distance has a minimum value in terms of the focal length.

    For a convex lens forming a real image of a real object, the distance between object and image cannot be less than $4f$. The minimum occurs when the object and image are each at $2f$, giving unit magnification.

    So the condition for an image to be obtainable at all is:

    $D\ge 4f$

    With the walls 3 m apart:

    $f\le \frac{D}{4}=\frac{3}{4}=0.75\text{m}$

    The maximum possible focal length is therefore 0.75 m, or 75 cm. A lens of longer focal length simply cannot bring the light to a focus within a 3 m span, because no pair of conjugate positions exists.

    ✦ Maximum focal length = 0.75 m

19. 193 marksNCERT Exercises, Chapter 9

    A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

    Hint. This is the displacement method; use the standard relation between the separation of the two lens positions and the object-screen distance.

    This is the displacement method. When the object-screen distance $D$ exceeds $4f$, there are two lens positions that throw a sharp image on the screen, and they are separated by a distance $d$ related to the focal length by:

    $f=\frac{D^2-d^2}{4D}$

    Substituting $D=90$ cm and $d=20$ cm:

    $f=\frac{90^2-20^2}{4\times 90}=\frac{8100-400}{360}=\frac{7700}{360}=21.4\text{cm}$

    The two positions exist because the lens formula is symmetric under interchanging object and image distances, so if $(u,v)$ works then so does $(v,u)$. This method is valued in the laboratory because it needs no knowledge of where the optical centre of the lens actually lies.

    ✦ f = 21.4 cm

20. 205 marksNCERT Exercises, Chapter 9

    (a) Determine the 'effective focal length' of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

    Hint. For (a), trace a parallel beam through both lenses in each direction and compare where it finally converges. For (b), take the image from the first lens as the object for the second.

    The lenses are $f_1=+30$ cm (convex) and $f_2=-20$ cm (concave), separated by $d=8.0$ cm.

    (a) Trace a parallel beam entering from the convex side. It converges towards $v_1=+30$ cm from the first lens, which is $30-8=22$ cm beyond the second lens, so it acts as a virtual object with $u_2=+22$ cm:

    $\frac{1}{v_2}=\frac{1}{-20}+\frac{1}{22}=\frac{-11+10}{220}=-\frac{1}{220}⟹v_2=-220\text{cm}$

    Now enter from the concave side. Parallel light gives $v_1=-20$ cm, which lies $20+8=28$ cm before the second lens, so $u_2=-28$ cm:

    $\frac{1}{v_2}=\frac{1}{30}-\frac{1}{28}=\frac{28-30}{840}=-\frac{1}{420}⟹v_2=-420\text{cm}$

    The two answers differ, $-220$ cm against $-420$ cm, so the answer does depend on which side the light enters. For comparison, the standard combination formula

    $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2}=\frac{1}{30}-\frac{1}{20}+\frac{8}{600}=-\frac{1}{300}$

    gives $F=-300$ cm regardless of direction. Since that single number predicts neither measured convergence point, the notion of an 'effective focal length' is not useful for this system: a separated pair of lenses needs its principal planes specified as well, and tracing the rays is the reliable route.

    (b) The object is 40 cm from the convex lens, so $u_1=-40$ cm:

    $\frac{1}{v_1}=\frac{1}{30}-\frac{1}{40}=\frac{4-3}{120}⟹v_1=+120\text{cm},m_1=\frac{v_1}{u_1}=-3$

    This image is $120-8=112$ cm beyond the concave lens, so $u_2=+112$ cm:

    $\frac{1}{v_2}=-\frac{1}{20}+\frac{1}{112}=-0.04107⟹v_2=-24.4\text{cm},m_2=\frac{v_2}{u_2}=-0.217$

    The total magnification is the product:

    $m=m_1{m}_2=(-3)(-0.217)=0.652$

    The image size is $1.5\times 0.652=0.98$ cm.

    ✦ (a) -220 cm from one side and -420 cm from the other, so the answer depends on the side and the notion is not useful (b) m = 0.652, image size about 0.98 cm

21. 214 marksNCERT Exercises, Chapter 9

    At what angle should a ray of light be incident on the face of a prism of refracting angle $60^{\circ }$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

    Hint. Total internal reflection at the second face means the angle there equals the critical angle; use the prism relation between the two internal angles.

    Total internal reflection at the second face occurs when the ray meets it exactly at the critical angle:

    $\sin C=\frac{1}{n}=\frac{1}{1.524}=0.6562⟹C=41.0^{\circ }$

    So the second internal angle is $r_2=41.0^{\circ }$. For any prism the two internal angles sum to the refracting angle:

    $r_1+r_2=A⟹r_1=60^{\circ }-41.0^{\circ }=19.0^{\circ }$

    Applying Snell's law at the first face:

    $\sin i=n\sin r_1=1.524\times \sin 19.0^{\circ }=1.524\times 0.3256=0.4962$

    $i=29.75^{\circ }$

    So light must strike the first face at about $29.75^{\circ }$. Any smaller angle of incidence makes $r_1$ smaller and hence $r_2$ larger, which pushes the second face further past the critical angle and still gives total internal reflection.

    ✦ i = 29.75 degrees, approximately 30 degrees

22. 224 marksNCERT Exercises, Chapter 9

    A card sheet divided into squares each of size 1 mm${}^2$ is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification (magnifying power) of the lens? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

    Hint. Note where the object sits relative to the focal length before computing anything, since that fixes the image position.

    (a) The card is at $u=-9$ cm and the focal length is $f=+9$ cm, so the object sits exactly at the focal point. The lens formula gives:

    $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{9}-\frac{1}{9}=0⟹v\to \infty$

    The image is formed at infinity. The linear magnification $m=v/u$ therefore grows without bound, and no finite area can be assigned to a square in the image. This is not a defect in the arithmetic: an image at infinity has no finite size, and the question is set up precisely so that this happens.

    (b) The angular magnification is still perfectly finite, because it compares the angle the image subtends with the angle the object would subtend at the near point:

    $\text{MP}=\frac{D}{f}=\frac{25}{9}=2.8$

    (c) They are not equal, and the two quantities are not even the same kind of thing. Linear magnification compares the physical size of image and object, and it depends on where the image forms, diverging when the image goes to infinity. Angular magnification compares the angle subtended at the eye with and without the instrument, and it stays finite because the rays leaving the lens are parallel and enter the eye at a definite angle. What a magnifier actually gives the eye is a larger angle, not a larger object.

    ✦ (a) The object is at the focus, so the image is at infinity and the linear magnification is unbounded (b) MP = 2.8 (c) No, since angular magnification compares angles at the eye while linear magnification compares sizes

23. 234 marksNCERT Exercises, Chapter 9

    (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case? Explain.

    Hint. Maximum magnifying power occurs when the virtual image is formed at the near point, not at infinity.

    (a) The magnifying power of a simple magnifier is greatest when the virtual image is formed at the near point, $v=-25$ cm, rather than at infinity. With $f=9$ cm:

    $\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=-\frac{1}{25}-\frac{1}{9}=-\frac{9+25}{225}=-\frac{34}{225}$

    $u=-6.6\text{cm}$

    The lens should be held about 6.6 cm from the card.

    (b) The linear magnification is:

    $m=\frac{v}{u}=\frac{-25}{-6.6}=3.8$

    (c) Here the two are numerically equal, since the magnifying power with the image at the near point is:

    $\text{MP}=1+\frac{D}{f}=1+\frac{25}{9}=3.8$

    They agree because the image is now formed at exactly the distance $D$ at which the unaided eye would view the object, so comparing sizes and comparing angles amount to the same calculation. The agreement is a special feature of this configuration and does not hold in general, as Exercise 9.22 showed.

    ✦ (a) 6.6 cm from the card (b) m = 3.8 (c) Yes, both equal 3.8, because the image is formed at the near point

24. 243 marksNCERT Exercises, Chapter 9

    What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm${}^2$? Would you be able to see the squares distinctly with your eyes very close to the magnifier?

    Hint. Convert the areal magnification into a linear magnification first, then find the object distance that produces it.

    Each square has area 1 mm${}^2$ in the object and is to have 6.25 mm${}^2$ in the image, so the areal magnification is 6.25 and the linear magnification is its square root:

    $m=\sqrt{6.25}=2.5$

    For a virtual image, $v=-2.5|u|$. Writing $u=-x$ so that $v=-2.5x$, the lens formula gives:

    $\frac{1}{-2.5x}-\frac{1}{-x}=\frac{1}{9}⟹\frac{1}{x}\left(1-0.4\right)=\frac{1}{9}$

    $x=0.6\times 9=5.4\text{cm}$

    So the card should be 5.4 cm from the lens, and the image forms at:

    $v=-2.5\times 5.4=-13.5\text{cm}$

    No, the squares could not be seen distinctly. The image lies only 13.5 cm from the eye, which is nearer than the near point of 25 cm. The eye cannot accommodate to focus that close, so although the magnification is larger, the image would appear blurred.

    ✦ Distance = 5.4 cm, with the image at 13.5 cm; no, since 13.5 cm is closer than the near point of 25 cm

25. 255 marksNCERT Exercises, Chapter 9

    Answer the following questions: (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? (b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back? (c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths? (e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

    Hint. Each part turns on distinguishing the angle subtended from where the object can be placed, and on where the rays actually cross after the eyepiece.

    (a) The comparison is not between object and image at the same place. Without the magnifier the object cannot be brought closer than the near point, 25 cm, so the angle it subtends is limited to $h/25$. The magnifier lets the object sit as close as its focal length while still producing an image the eye can focus, so the angle becomes roughly $h/f$. The gain is in being able to bring the object much closer, not in the image subtending more than the object at the same distance.

    (b) Yes, it decreases slightly. Moving the eye back increases its distance from the image, so the angle the image subtends at the eye falls. The effect is small while the image is far away, and it becomes negligible when the image is at infinity, but the maximum angular magnification is obtained with the eye close to the lens.

    (c) Two limits bite. Shortening the focal length means increasing the curvature of the surfaces, which rapidly worsens aberrations, both spherical and chromatic, so the image degrades faster than it magnifies. It also makes the lens very small and the working distance impractically short. In practice a simple magnifier is limited to about $9\times$ before the image quality becomes unusable.

    (d) The magnifying power of a compound microscope is roughly $\frac{L}{f_o}\cdot \frac{D}{f_e}$, so it is inversely proportional to both focal lengths. Making $f_o$ short gives a large linear magnification at the objective, and making $f_e$ short gives a large angular magnification at the eyepiece. Shortening either one alone leaves the other factor limiting the result.

    (e) All the rays leaving the eyepiece pass through a small region called the eye ring, the image of the objective formed by the eyepiece. Placing the eye there collects the greatest amount of light and gives the widest field of view. Sitting directly against the eyepiece lens puts the eye in front of that point, so the field of view is smaller. The eye ring typically lies a few centimetres beyond the eyepiece, and its exact position follows from treating the objective as an object for the eyepiece.

    ✦ (a) The gain comes from being able to place the object much closer than the near point (b) Yes, it decreases slightly (c) Aberrations and impractical working distances (d) Magnifying power varies inversely with both focal lengths (e) The eye ring collects all the emergent light and gives the widest field of view

26. 264 marksNCERT Exercises, Chapter 9

    An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

    Hint. Split the required total magnification between eyepiece and objective, then find the object and image distances that deliver the objective's share.

    Take the final image at the near point, $D=25$ cm. The eyepiece then contributes:

    $m_e=1+\frac{D}{f_e}=1+\frac{25}{5}=6$

    The objective must supply the rest:

    $m_o=\frac{M}{m_e}=\frac{30}{6}=5$

    Since $m_o=v_o/u_o=5$ in magnitude, write $u_o=-x$ and $v_o=5x$. The lens formula for the objective gives:

    $\frac{1}{5x}+\frac{1}{x}=\frac{1}{f_o}⟹\frac{1.2}{x}=\frac{1}{1.25}$

    $x=1.5\text{cm},v_o=7.5\text{cm}$

    For the eyepiece with $v_e=-25$ cm:

    $\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{5}=-\frac{6}{25}⟹u_e=-4.17\text{cm}$

    The separation between the lenses is:

    $L=v_o+|u_e|=7.5+4.17=11.67\text{cm}$

    So place the object 1.5 cm from the objective and set the lenses 11.67 cm apart.

    ✦ Object 1.5 cm from the objective, with the lenses separated by 11.67 cm

27. 273 marksNCERT Exercises, Chapter 9

    A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25 cm)?

    Hint. Normal adjustment uses the plain ratio of focal lengths; the near-point setting carries an extra factor from the eyepiece.

    (a) In normal adjustment the final image is at infinity and the magnifying power is simply the ratio of the focal lengths:

    $M=\frac{f_o}{f_e}=\frac{140}{5.0}=28$

    (b) With the final image at the near point, the eyepiece works harder and contributes an extra factor:

    $M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=28\left(1+\frac{5}{25}\right)=28\times 1.2=33.6$

    The near-point setting gives about 20 per cent more magnification, but it requires the eye to accommodate continuously, which is tiring. Normal adjustment is preferred for extended observation because the eye stays relaxed.

    ✦ (a) M = 28 (b) M = 33.6

28. 284 marksNCERT Exercises, Chapter 9

    (a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and the eyepiece? (b) 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? (c) What is the height of the final image of the tower if it is formed at 25 cm?

    Hint. The objective forms its image at its focal plane; scale that image by the eyepiece's magnification for the final size.

    (a) In normal adjustment the separation is the sum of the focal lengths:

    $L=f_o+f_e=140+5.0=145\text{cm}$

    (b) The tower subtends an angle at the objective of:

    $\theta =\frac{100}{3000}=0.0333\text{rad}$

    The objective forms its image in its focal plane, so the image height is:

    $h=f_o\theta =140\times 0.0333=4.67\text{cm}$

    (c) The eyepiece then magnifies this intermediate image. With the final image at 25 cm, its angular magnification is:

    $m_e=1+\frac{D}{f_e}=1+\frac{25}{5}=6$

    so the final image height is:

    $h^{\prime }=4.67\times 6=28\text{cm}$

    ✦ (a) 145 cm (b) 4.67 cm (c) 28 cm

29. 294 marksNCERT Exercises, Chapter 9

    A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

    Hint. Find where the primary alone would form the image, then treat that point as a virtual object for the secondary mirror.

    The primary is a concave mirror of focal length:

    $f_1=\frac{R_1}{2}=\frac{220}{2}=110\text{mm}$

    For an object at infinity it would form an image at its focus, 110 mm from the primary. The secondary sits 20 mm from the primary, so that image lies:

    $110-20=90\text{mm}$

    beyond the secondary. The converging light is intercepted before reaching it, so this acts as a virtual object for the secondary.

    The secondary is convex with focal length:

    $f_2=\frac{R_2}{2}=\frac{140}{2}=70\text{mm}$

    Applying the mirror formula with $u=+90$ mm and $f=+70$ mm, both measured on the same side:

    $\frac{1}{v}=\frac{1}{70}-\frac{1}{90}=\frac{9-7}{630}=\frac{2}{630}$

    $v=315\text{mm}$

    The final image forms 315 mm from the secondary mirror, on the far side, where it emerges through the hole in the primary and can be examined with an eyepiece.

    ✦ The final image is 315 mm from the small mirror

30. 303 marksNCERT Exercises, Chapter 9

    Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.29. A current in the coil produces a deflection of $3.5^{\circ }$ of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

    Hint. When a mirror turns through an angle, the reflected ray turns through twice that angle.

    Rotating a mirror through an angle $\theta$ turns the reflected ray through $2\theta$, because both the incident angle and the reflected angle change by $\theta$.

    So the reflected beam swings through:

    $2\times 3.5^{\circ }=7.0^{\circ }$

    On a screen at distance $d=1.5$ m, the spot moves through:

    $x=d\tan (7.0^{\circ })=1.5\times 0.1228=0.184\text{m}$

    The displacement is about 18.4 cm. This doubling is what makes the lamp-and-scale arrangement so sensitive, since a small rotation of the coil produces a conveniently large movement of the spot.

    ✦ Displacement = 0.184 m, about 18.4 cm

31. 315 marksNCERT Exercises, Chapter 9

    Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

    Hint. When the image coincides with the object, the light must strike the mirror normally, so the needle sits at the focal point of the lens-liquid combination.

    When the inverted image coincides with the needle, the rays must fall normally on the plane mirror and retrace their path. That happens only when the needle is at the focal point of the lens system above the mirror.

    So the two measurements give focal lengths directly:

    $F=45.0\text{cm}\text{(lens plus liquid)},f_1=30.0\text{cm}\text{(lens alone)}$

    The liquid layer forms a second lens in contact, so:

    $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}⟹\frac{1}{f_2}=\frac{1}{45}-\frac{1}{30}=\frac{2-3}{90}=-\frac{1}{90}$

    $f_2=-90\text{cm}$

    First find the radius of the glass lens. For an equiconvex lens with $R_1=+R$ and $R_2=-R$:

    $\frac{1}{f_1}=(n_g-1)\frac{2}{R}⟹R=2f_1(n_g-1)=2\times 30\times 0.50=30\text{cm}$

    The liquid between the curved lens and the flat mirror forms a plano-concave lens, with its upper surface matching the lens at $R_1=-30$ cm and its lower surface flat, $R_2=\infty$:

    $\frac{1}{f_2}=(n_l-1)\left(\frac{1}{-30}-0\right)$

    $-\frac{1}{90}=-\frac{n_l-1}{30}⟹n_l-1=\frac{30}{90}=0.333$

    $n_l=1.33$

    The liquid is water, and the method is a standard laboratory way of measuring a refractive index without a prism.

    ✦ n_liquid = 1.33

Solutions written by the tuition.in editorial team and checked against leph201.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.