NCERT Class 12 Physics — Ray Optics and Optical Instruments, Chapter 9 Exercises. All 31 questions solved.

The exercises at the end of Chapter 9 cover the whole chapter. Questions 9.1, 9.2, 9.15 and 9.29 are about spherical mirrors; 9.3 to 9.6, 9.16, 9.17 and 9.21 about refraction at flat surfaces, total internal reflection and the prism; 9.7 to 9.10, 9.18 to 9.20 and 9.31 about thin lenses; and 9.11 to 9.14 and 9.22 to 9.28 about the magnifying glass, the compound microscope and the telescope. Almost everything runs on three sets of results:

• Mirror and lens equations — for a mirror $\dfrac1v + \dfrac1u = \dfrac1f$ with $m = -\dfrac vu$; for a thin lens $\dfrac1v – \dfrac1u = \dfrac1f$ with $m = \dfrac vu$.
• Snell’s law — $n_1\sin i = n_2\sin r$, with the critical angle for a denser-to-rarer surface given by $\sin i_c = n_2/n_1$.
• Magnifying powers — a magnifying glass gives $m = 1 + D/f$ with the image at the near point $D = 25\ \text{cm}$; a compound microscope gives $m = m_o m_e$; a telescope in normal adjustment gives $m = f_o/f_e$.

Every answer uses the Cartesian sign convention, as the chapter does. All distances are measured from the pole of the mirror or the optical centre of the lens. Distances in the direction of the incident light are positive and distances against it are negative; heights above the axis are positive. So with light travelling from left to right, a real object has $u < 0$, a concave mirror has $f < 0$, a convex mirror $f > 0$, a convex lens $f > 0$ and a concave lens $f < 0$.

Key insight. Put the signs in once, at the start, and then trust the equation. A negative $v$ from a mirror means a real image in front of it; a negative $v$ from a lens means a virtual image on the object’s side; a positive $u$ means a virtual object, light that was converging before it arrived (9.8, 9.20, 9.29). Nearly every wrong answer in this exercise comes from deciding a sign by intuition half-way through a calculation.

Question 9.1

A small candle, $2.5\ \text{cm}$ in size, is placed at $27\ \text{cm}$ in front of a concave mirror of radius of curvature $36\ \text{cm}$. 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?

Solution. The screen must go where the real image forms, so we need $v$. A concave mirror has $f = R/2$, and with the incident light travelling towards the mirror both $f$ and $u$ are measured against it: $f = -18\ \text{cm}$, $u = -27\ \text{cm}$. The mirror equation gives

$$\frac1v = \frac1f – \frac1u = -\frac1{18} + \frac1{27} = \frac{-3 + 2}{54} = -\frac1{54}, \qquad v = -54\ \text{cm}$$

The negative sign puts the image on the same side as the candle, in front of the mirror, so it is real and can be caught on a screen $54\ \text{cm}$ from the mirror. The magnification is

$$m = -\frac vu = -\frac{-54}{-27} = -2, \qquad h’ = mh = -2\times2.5 = -5.0\ \text{cm}$$

so the image is inverted and twice the size of the candle. This is what we expect: the candle is between $F$ and $C$, and an object there always gives a real, magnified image beyond $C$.

As the candle moves towards $F$, $1/u$ moves towards $1/f$ and $1/v$ towards zero, so the image moves further away and the screen has to be moved back. At $u = f$ the image goes to infinity, and once the candle is inside $F$ the image is virtual, behind the mirror, and cannot be caught on a screen at all.

Question 9.2

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

Solution. For a convex mirror the focus lies behind the mirror, in the direction of the incident light, so $f = +15\ \text{cm}$; the needle is in front, so $u = -12\ \text{cm}$.

$$\frac1v = \frac1f – \frac1u = \frac1{15} + \frac1{12} = \frac{4 + 5}{60} = \frac{9}{60}, \qquad v = +6.7\ \text{cm}$$

A positive $v$ places the image behind the mirror, so it is virtual.

$$m = -\frac vu = -\frac{20/3}{-12} = \frac59 \approx 0.56, \qquad h’ = \frac59\times4.5 = 2.5\ \text{cm}$$

$m$ is positive and less than one, so the image is erect and diminished. As the needle moves away, $1/u \to 0$ and $1/v \to 1/f$: the image moves back towards the focus but never passes it, and $m = -v/u$ shrinks towards zero.

Question 9.3

A tank is filled with water to a height of $12.5\ \text{cm}$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be $9.4\ \text{cm}$. 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?

Solution. Viewed from almost directly above, rays leaving the water bend away from the normal, so the needle appears raised. For near-normal viewing the refractive index is real depth over apparent depth:

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

In the new liquid the same needle appears at a depth of $12.5/1.63 = 7.67\ \text{cm}$. The image of the needle has risen by $9.4 – 7.67 = 1.73\ \text{cm}$, closer to the microscope, so the microscope must be raised by the same amount to keep its focus on it.

Question 9.4

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)].

Solution. Each of the first two figures gives one refractive index relative to air, and the water-glass surface needs the index of glass relative to water, which is their ratio. From Fig. 9.27(a), with the ray in air at $60^\circ$ and in glass at $35^\circ$:

$$n_{ga} = \frac{\sin60^\circ}{\sin35^\circ} = \frac{0.8660}{0.5736} = 1.51$$

From Fig. 9.27(b), with the ray in air at $60^\circ$ and in water at $41^\circ$ (the book’s figure prints $47^\circ$; the note below explains why $41^\circ$ is used):

$$n_{wa} = \frac{\sin60^\circ}{\sin41^\circ} = \frac{0.8660}{0.6561} = 1.32$$

Then $n_{gw} = n_{ga}/n_{wa} = 1.51/1.32 = 1.144$, and Snell’s law at the water-glass surface gives

$$\sin r = \frac{\sin45^\circ}{n_{gw}} = \frac{0.7071}{1.144} = 0.618, \qquad r \approx 38^\circ$$

The ray bends towards the normal on entering glass, as it must, since glass is optically denser than water.

Why $41^\circ$, and not the $47^\circ$ printed in the figure. Read literally, Fig. 9.27(b) would give water a refractive index of $\sin60^\circ/\sin47^\circ = 1.18$. Water’s refractive index is $1.33$, and a ray entering water from air at $60^\circ$ really refracts at $\sin^{-1}(\sin60^\circ/1.33) = 40.6^\circ$, about $41^\circ$. The book’s own answer key works with $n_{wa} = 1.32$, which is exactly what $41^\circ$ gives, and arrives at $38^\circ$. So the $47^\circ$ in the figure is a misprint, and this solution uses $41^\circ$, in agreement with both the key and real water. Taken at face value, the printed $47^\circ$ would lead to about $34^\circ$ — but only for a liquid of refractive index $1.18$, which is not water.

Question 9.5

A small bulb is placed at the bottom of a tank containing water to a depth of $80\ \text{cm}$. 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.)

Solution. Light from the bulb reaches the surface at every angle, but only rays meeting it at less than the critical angle get out; the rest are totally reflected back into the water. The rays that just escape meet the surface at $i_c$, and they mark out a circle directly above the bulb. The critical angle for water to air is

$$\sin i_c = \frac1{1.33} = 0.752, \qquad i_c = 48.75^\circ, \qquad \tan i_c = 1.140$$

A ray from the bulb at $i_c$ to the vertical reaches the surface a horizontal distance $r = 0.8\tan i_c$ from the point directly above it:

$$r = 0.8\times1.140 = 0.912\ \text{m}, \qquad A = \pi r^2 = \pi(0.912)^2 = 2.6\ \text{m}^2$$

Question 9.6

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.

Solution. At minimum deviation the ray passes symmetrically through the prism, and the prism formula relates the refractive index to the angle $A$ and the deviation $D_m$:

$$n = \frac{\sin\left(\dfrac{A + D_m}{2}\right)}{\sin\left(\dfrac A2\right)} = \frac{\sin50^\circ}{\sin30^\circ} = \frac{0.7660}{0.5} = 1.532$$

In water, what bends the light is the index of the glass relative to water, so the same formula applies with $n$ replaced by $n_{gw} = 1.532/1.33 = 1.152$:

$$\sin\left(\frac{60^\circ + D_m’}{2}\right) = 1.152\times\sin30^\circ = 0.576, \qquad \frac{60^\circ + D_m’}{2} = 35.2^\circ, \qquad D_m’ = 10.3^\circ$$

The deviation falls sharply because glass and water differ far less in refractive index than glass and air.

Question 9.7

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 $20\ \text{cm}$?

Solution. The lens maker’s formula links the focal length to the glass and the shape of the surfaces: $\dfrac1f = (n – 1)\left(\dfrac1{R_1} – \dfrac1{R_2}\right)$. For a double-convex lens with equal faces, light meets the first face with its centre of curvature ahead ($R_1 = +R$) and the second with its centre behind ($R_2 = -R$), so the bracket is $2/R$:

$$\frac1{20} = (1.55 – 1)\frac2R \quad\Rightarrow\quad R = 2\times0.55\times20 = 22\ \text{cm}$$

Question 9.8

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

Solution. The lens receives light that is already converging towards $P$, $12\ \text{cm}$ beyond it. For the lens this is a virtual object: the rays head towards a point on the far side, the side the light travels towards, so $u = +12\ \text{cm}$.

Question 9.8 (a)

Solution. With $f = +20\ \text{cm}$:

$$\frac1v = \frac1f + \frac1u = \frac1{20} + \frac1{12} = \frac{3 + 5}{60} = \frac8{60}, \qquad v = +7.5\ \text{cm}$$

The convex lens adds its own convergence to the beam’s, so the rays meet sooner, $7.5\ \text{cm}$ beyond the lens instead of $12\ \text{cm}$. A positive $v$ means a real image on the far side.

Question 9.8 (b)

Solution. With $f = -16\ \text{cm}$:

$$\frac1v = \frac1f + \frac1u = -\frac1{16} + \frac1{12} = \frac{-3 + 4}{48} = \frac1{48}, \qquad v = +48\ \text{cm}$$

The concave lens takes away part of the convergence but not all of it, because its power ($1/16$) is less than the beam’s ($1/12$). The rays still meet, but much further away.

Question 9.9

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

Solution. For a concave lens $f = -21\ \text{cm}$, and the object in front gives $u = -14\ \text{cm}$:

$$\frac1v = \frac1f + \frac1u = -\frac1{21} – \frac1{14} = -\frac{2 + 3}{42} = -\frac5{42}, \qquad v = -8.4\ \text{cm}$$

A negative $v$ puts the image on the same side as the object, so it is virtual.

$$m = \frac vu = \frac{-8.4}{-14} = 0.6, \qquad h’ = 0.6\times3.0 = 1.8\ \text{cm}$$

As the object moves away, $1/u \to 0$ and $v \to f$: the image moves towards the focus, $21\ \text{cm}$ from the lens, but never reaches it, and it shrinks. There is nothing special about the object being at the focus of a concave lens: with $u = -21\ \text{cm}$, $1/v = -2/21$ and the image is at $10.5\ \text{cm}$, not at infinity.

Question 9.10

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

Solution. For thin lenses in contact the powers add, so $\dfrac1F = \dfrac1{f_1} + \dfrac1{f_2}$, with the signs of the convention:

$$\frac1F = \frac1{30} – \frac1{20} = \frac{2 – 3}{60} = -\frac1{60}, \qquad F = -60\ \text{cm}$$

The concave lens is the stronger of the two, so it wins and the pair diverges light.

Question 9.11

A compound microscope consists of an objective lens of focal length $2.0\ \text{cm}$ and an eyepiece of focal length $6.25\ \text{cm}$ separated by a distance of $15\ \text{cm}$. 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 ($25\ \text{cm}$), and (b) at infinity? What is the magnifying power of the microscope in each case?

Solution. Work backwards from the eye. The final image fixes where the eyepiece’s object must be; that object is the image formed by the objective, which fixes $v_o$; and the lens equation for the objective then gives $u_o$. The magnifying power is the objective’s magnification $v_o/|u_o|$ times the eyepiece’s angular magnification.

Question 9.11 (a)

Solution. The final image is virtual and on the object side of the eyepiece, so $v_e = -25\ \text{cm}$:

$$\frac1{u_e} = \frac1{v_e} – \frac1{f_e} = -\frac1{25} – \frac1{6.25} = -0.20, \qquad u_e = -5.0\ \text{cm}$$

The intermediate image is $5.0\ \text{cm}$ in front of the eyepiece, so it is $v_o = 15 – 5 = 10\ \text{cm}$ beyond the objective. For the objective,

$$\frac1{u_o} = \frac1{v_o} – \frac1{f_o} = \frac1{10} – \frac12 = -0.40, \qquad u_o = -2.5\ \text{cm}$$

With the final image at the near point the eyepiece contributes $m_e = 1 + D/f_e$:

$$m = \frac{v_o}{|u_o|}\left(1 + \frac D{f_e}\right) = \frac{10}{2.5}\left(1 + \frac{25}{6.25}\right) = 4\times5 = 20$$

Question 9.11 (b)

Solution. For a final image at infinity the intermediate image must sit at the focus of the eyepiece, $u_e = -6.25\ \text{cm}$, so $v_o = 15 – 6.25 = 8.75\ \text{cm}$:

$$\frac1{u_o} = \frac1{8.75} – \frac12 = -0.386, \qquad u_o = -2.59\ \text{cm}$$

Now the eyepiece gives $m_e = D/f_e$:

$$m = \frac{v_o}{|u_o|}\cdot\frac D{f_e} = \frac{8.75}{2.59}\times\frac{25}{6.25} = 3.375\times4 = 13.5$$

Question 9.12

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

Solution. “Sharp focus” for a person with a normal near point means the final image is at $25\ \text{cm}$. Working in centimetres, the objective has $f_o = 0.8\ \text{cm}$ and $u_o = -0.9\ \text{cm}$:

$$\frac1{v_o} = \frac1{f_o} + \frac1{u_o} = \frac1{0.8} – \frac1{0.9} = 0.1389, \qquad v_o = 7.2\ \text{cm}$$

For the eyepiece, $v_e = -25\ \text{cm}$ and $f_e = 2.5\ \text{cm}$:

$$\frac1{u_e} = -\frac1{25} – \frac1{2.5} = -0.44, \qquad |u_e| = \frac{25}{11} = 2.27\ \text{cm}$$

The separation is $v_o + |u_e| = 7.2 + 2.27 = 9.47\ \text{cm}$, and

$$m = \frac{v_o}{|u_o|}\left(1 + \frac D{f_e}\right) = \frac{7.2}{0.9}\left(1 + \frac{25}{2.5}\right) = 8\times11 = 88$$

Question 9.13

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

Solution. In normal adjustment the objective forms the image of a distant object at its focal point, and the eyepiece is placed so that this point is also its own focus, sending the final image to infinity. Then $m = f_o/f_e$ and the lenses are $f_o + f_e$ apart:

$$m = \frac{144}{6.0} = 24, \qquad L = 144 + 6.0 = 150\ \text{cm}$$

Question 9.14

(a) A giant refracting telescope at an observatory has an objective lens of focal length $15\ \text{m}$. If an eyepiece of focal length $1.0\ \text{cm}$ 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 $3.48\times10^6\ \text{m}$, and the radius of lunar orbit is $3.8\times10^8\ \text{m}$.

Question 9.14 (a)

Solution. Both focal lengths must be in the same unit: $f_o = 1500\ \text{cm}$.

$$m = \frac{f_o}{f_e} = \frac{1500}{1.0} = 1500$$

Question 9.14 (b)

Solution. The moon is so far away that its image forms at the focal plane of the objective. A ray through the centre of the lens is undeviated, so the image subtends the same angle at the lens as the moon does:

$$\alpha = \frac{3.48\times10^6}{3.8\times10^8} = 9.16\times10^{-3}\ \text{rad}, \qquad d = f_o\,\alpha = 15\times9.16\times10^{-3} = 0.137\ \text{m}$$

Question 9.15

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. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Solution. In every part the object is real and in front of the mirror, so $u < 0$. Write the mirror equation as $\dfrac1v = \dfrac1f – \dfrac1u$ and read off the sign and size of $v$; a negative $v$ is a real image in front of the mirror and a positive $v$ a virtual one behind it, and $|m| = |v|/|u|$.

Question 9.15 (a)

Solution. For a concave mirror $f < 0$. “Between $f$ and $2f$” means $2f < u < f$, all negative, so $\dfrac1f < \dfrac1u < \dfrac1{2f} < 0$. Subtracting,

$$\frac1{2f} = \frac1f – \frac1{2f} < \frac1f – \frac1u = \frac1v < 0$$

So $1/v$ is negative, and the image is real; and $1/v$ lies between $1/2f$ and zero, so $|v| > 2|f|$, which means $v < 2f$.

Question 9.15 (b)

Solution. For a convex mirror $f > 0$, and $u < 0$ makes $-1/u$ positive. So

$$\frac1v = \frac1f – \frac1u > 0$$

for every position of the object, and $v$ is always positive: the image is always behind the mirror.

Question 9.15 (c)

Solution. With $f > 0$ and $u < 0$, both terms in $\dfrac1v = \dfrac1f + \dfrac1{|u|}$ are positive. That makes $\dfrac1v > \dfrac1f$, so $0 < v < f$: the image lies between the pole and the focus. It also makes $\dfrac1v > \dfrac1{|u|}$, so $v < |u|$ and $m = -\dfrac vu = \dfrac v{|u|} < 1$.

Question 9.15 (d)

Solution. For a concave mirror $f < 0$, and the object between the pole and the focus means $f < u < 0$, so $|u| < |f|$. Then

$$\frac1v = -\frac1{|f|} + \frac1{|u|} > 0$$

because $1/|u| > 1/|f|$; so $v > 0$ and the image is virtual. Also $\dfrac1v = \dfrac1{|u|} – \dfrac1{|f|} < \dfrac1{|u|}$, so $v > |u|$ and $m = v/|u| > 1$.

Question 9.16

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

Solution. The slab makes the part of the path that lies in glass look shorter by the factor $1/n$, as in apparent depth, while the parts in air are unchanged. So the whole shift comes from the glass and equals

$$\text{shift} = t\left(1 – \frac1n\right) = 15\left(1 – \frac1{1.5}\right) = 5.0\ \text{cm}$$

Only the thickness of glass appears in this, not where the slab is, so for near-normal viewing the pin appears raised by the same amount wherever the slab is held between the pin and the eye. (The $50\ \text{cm}$ is not needed.)

Question 9.17

(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, as shown in the figure. (b) What is the answer if there is no outer covering of the pipe?

Question 9.17 (a)

Solution. Total reflection at the wall needs the angle $i’$ there to exceed the critical angle for fibre to coating:

$$\sin i_c’ = \frac{1.44}{1.68} = 0.857, \qquad i_c’ = 59^\circ$$

The ray inside the fibre makes angle $r$ with the axis, and the wall is parallel to the axis, so $i’ = 90^\circ – r$. The condition $i’ > 59^\circ$ therefore means $r < 31^\circ$. At the end face the ray refracts from air into the fibre, $\sin i = 1.68\sin r$, so the largest allowed angle of incidence is

$$\sin i_{\max} = 1.68\sin31^\circ = 0.865, \qquad i_{\max} \approx 60^\circ$$

Any ray entering at a smaller angle to the axis has a smaller $r$ and a larger $i’$, so it is also totally reflected.

Question 9.17 (b)

Solution. Without a coating the wall is a fibre-air surface, and the critical angle falls to

$$\sin i_c’ = \frac1{1.68}, \qquad i_c’ = 36.5^\circ$$

The worst case is a ray grazing the end face, $i = 90^\circ$. It refracts to $\sin r = 1/1.68$, so $r = 36.5^\circ$ and it meets the wall at $i’ = 90^\circ – 36.5^\circ = 53.5^\circ$, which is still more than $36.5^\circ$. Every other ray has a smaller $r$ and a larger $i’$. So every ray that enters the end face is totally reflected, and $i’$ always lies between $53.5^\circ$ and $90^\circ$.

Note on the book’s answer. The key writes “all incident rays (in the range $53.5^\circ < i < 90^\circ$)”. The range printed is that of $i’$, the angle at the wall, not of the angle of incidence $i$, which can be anything from $0$ to $90^\circ$; the key’s own words “all incident rays” agree with the answer here.

Question 9.18

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

Solution. The object and the screen are a fixed distance $s = 3\ \text{m}$ apart, and the lens goes somewhere between them. If the bulb is a distance $x$ from the lens, the image must be $s – x$ beyond it, and the lens equation (with $u = -x$, $v = s – x$) becomes

$$\frac1{s – x} + \frac1x = \frac1f \quad\Rightarrow\quad x^2 – sx + sf = 0$$

A lens position exists only if this quadratic has a real root, which needs $s^2 – 4sf \ge 0$, that is $f \le s/4$. A lens with a longer focal length cannot throw a real image as close as $3\ \text{m}$ from the object, wherever it is put.

$$f_{\max} = \frac s4 = \frac{3}{4} = 0.75\ \text{m}$$

Question 9.19

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

Solution. Light paths are reversible, so if the lens gives a sharp image with the object $x$ from it and the screen $y$ from it, it also gives one with the two distances swapped. The two positions of the lens are these two arrangements. So $x + y = 90$, and moving the lens from one position to the other changes its distance from the object by $y – x = 20$. Solving, $x = 35\ \text{cm}$ and $y = 55\ \text{cm}$. With $u = -35\ \text{cm}$ and $v = +55\ \text{cm}$:

$$\frac1f = \frac1v – \frac1u = \frac1{55} + \frac1{35} = \frac{90}{1925}, \qquad f = 21.4\ \text{cm}$$

This is the general result $f = \dfrac{D^2 – d^2}{4D} = \dfrac{90^2 – 20^2}{4\times90}$ for object-screen distance $D$ and lens displacement $d$.

Question 9.20

(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed $8.0\ \text{cm}$ 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 $1.5\ \text{cm}$ 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 $40\ \text{cm}$. Determine the magnification produced by the two-lens system, and the size of the image.

Question 9.20 (a)

Solution. Separated lenses have to be taken one at a time: the image formed by the first lens is the object for the second. Send a parallel beam in from each side in turn.

Parallel light enters the convex lens first. It focuses at $v_1 = +30\ \text{cm}$. The concave lens is $8\ \text{cm}$ from the convex one, so the beam reaches it still converging, towards a point $22\ \text{cm}$ beyond it: a virtual object, $u_2 = +22\ \text{cm}$. With $f_2 = -20\ \text{cm}$,

$$\frac1{v_2} = \frac1{f_2} + \frac1{u_2} = -\frac1{20} + \frac1{22} = -\frac1{220}, \qquad v_2 = -220\ \text{cm}$$

The beam leaves diverging, as if from a point $220\ \text{cm}$ in front of the concave lens, which is $216\ \text{cm}$ from the centre of the two-lens system.

Parallel light enters the concave lens first. It diverges from $v_1 = -20\ \text{cm}$, a real object for the convex lens at $u_2 = -(20 + 8) = -28\ \text{cm}$. With $f_2 = +30\ \text{cm}$,

$$\frac1{v_2} = \frac1{30} – \frac1{28} = -\frac2{840}, \qquad v_2 = -420\ \text{cm}$$

This beam also leaves diverging, now from a point $420\ \text{cm}$ in front of the convex lens, $416\ \text{cm}$ from the centre of the system.

The two answers differ, so the apparent focal point depends on which side the light comes from. Nor is there a single lens equation of the form $1/v – 1/u = 1/F$ with one constant $F$ that works for every $u$ for this pair, since the result depends on $f_1$, $f_2$ and the separation in a more complicated way. A single “effective focal length” is therefore not a useful description of this system.

Question 9.20 (b)

Solution. The object is $40\ \text{cm}$ in front of the convex lens, $u_1 = -40\ \text{cm}$:

$$\frac1{v_1} = \frac1{30} – \frac1{40} = \frac1{120}, \qquad v_1 = 120\ \text{cm}, \qquad m_1 = \frac{v_1}{u_1} = \frac{120}{-40} = -3$$

That image would form $120 – 8 = 112\ \text{cm}$ beyond the concave lens, so it is a virtual object for it, $u_2 = +112\ \text{cm}$:

$$\frac1{v_2} = -\frac1{20} + \frac1{112} = -\frac{92}{2240}, \qquad v_2 = -\frac{112\times20}{92} = -24.3\ \text{cm}, \qquad m_2 = \frac{v_2}{u_2} = -\frac{20}{92}$$

The total magnification is the product:

$$m = m_1m_2 = (-3)\left(-\frac{20}{92}\right) = 0.652, \qquad h’ = 0.652\times1.5 = 0.98\ \text{cm}$$

The final image is virtual (it lies $24.3\ \text{cm}$ in front of the concave lens, between the object and the convex lens) and, since $m$ is positive, erect.

Question 9.21

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 $1.524$.

Solution. “Just suffers total internal reflection” means the ray meets the second face at exactly the critical angle:

$$\sin i_c = \frac1{1.524} = 0.656, \qquad i_c = 41.0^\circ$$

Inside a prism the angles of refraction at the two faces add up to the prism angle, $r_1 + r_2 = A$, because the two normals meet at an angle $A$. With $r_2 = i_c$, the ray must be refracted at the first face to $r_1 = 60^\circ – 41.0^\circ = 19.0^\circ$, and Snell’s law at that face gives the angle of incidence:

$$\sin i = 1.524\sin19.0^\circ = 0.496, \qquad i \approx 29.7^\circ \approx 30^\circ$$

At any smaller angle of incidence $r_1$ is smaller, $r_2$ is larger than $i_c$, and the ray is totally reflected at the second face.

Question 9.22

A card sheet divided into squares each of size $1\ \text{mm}^2$ is being viewed at a distance of $9\ \text{cm}$ through a magnifying glass (a converging lens of focal length $10\ \text{cm}$) 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.

A misprint in the question. The current reprint prints the focal length as $9\ \text{cm}$. That cannot be what is intended: with the card $9\ \text{cm}$ away it would sit exactly at the focus, the image would be at infinity and part (a) would have no answer. The book’s answer key works with $10\ \text{cm}$, as do Exercises 9.23 and 9.24, which reuse this lens, and earlier editions print $10\ \text{cm}$. The solution below uses $10\ \text{cm}$.

Question 9.22 (a)

Solution. The card is inside the focal length, so the lens forms a virtual, magnified image. With $u = -9\ \text{cm}$ and $f = +10\ \text{cm}$:

$$\frac1v = \frac1f + \frac1u = \frac1{10} – \frac19 = -\frac1{90}, \qquad v = -90\ \text{cm}, \qquad m = \frac vu = \frac{-90}{-9} = 10$$

Every length in the image is ten times the original, so every area is $10^2 = 100$ times as large.

Question 9.22 (b)

Solution. Magnifying power compares the angle the object subtends at the eye when viewed through the lens with the angle it would subtend at the near point, $25\ \text{cm}$, without the lens. With the lens close to the eye, the image subtends the same angle as the card itself at $9\ \text{cm}$, $h/9$, against $h/25$ at the near point:

$$m_{\text{ang}} = \frac{h/9}{h/25} = \frac{25}{9} = 2.8$$

Question 9.22 (c)

Solution. No. The magnification $|v/u|$ compares the size of the image with the size of the object. The magnifying power $25/|u|$ compares angles, and the angle is what sets how big something looks. The image here is ten times larger than the card but also ten times further away, so it subtends the same angle as the card; what the lens really achieves is to let the card be held at $9\ \text{cm}$ instead of $25\ \text{cm}$. The two numbers agree only when the image is at the near point, $|v| = 25\ \text{cm}$, which is not the case here.

Question 9.23

(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.

Question 9.23 (a)

Solution. The magnifying power $25/|u|$ grows as the card comes closer, but the image must stay far enough away to be seen distinctly. The closest the card can come is where the image is at the near point, $v = -25\ \text{cm}$:

$$\frac1u = \frac1v – \frac1f = -\frac1{25} – \frac1{10} = -0.14, \qquad u = -7.14\ \text{cm}$$

Question 9.23 (b)

Solution.

$$m = \frac vu = \frac{-25}{-7.14} = 3.5$$

Question 9.23 (c)

Solution. The magnifying power is $25/|u| = 25/7.14 = 3.5$, the same as the magnification. This is the one case where the two coincide: the image is at $25\ \text{cm}$, exactly where the object would be viewed without the lens, so comparing their angles is the same as comparing their sizes.

Question 9.24

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 $6.25\ \text{mm}^2$. Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note: Exercises 9.22 to 9.24 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Solution. Areas scale as the square of the magnification, so the linear magnification must be $m = \sqrt{6.25/1} = 2.5$. The image is virtual and erect, so $v = 2.5u$, and the lens equation gives

$$\frac1{2.5u} – \frac1u = \frac1{10} \quad\Rightarrow\quad \frac{1 – 2.5}{2.5u} = \frac1{10} \quad\Rightarrow\quad u = -6\ \text{cm}, \qquad v = -15\ \text{cm}$$

The image would be only $15\ \text{cm}$ from the eye, closer than the near point of $25\ \text{cm}$, so the eye cannot focus on it.

Question 9.25

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?

Question 9.25 (a)

Solution. The image is bigger than the object, but it is also further away, so at the eye it subtends the same angle as the object in its actual position. The gain comes from where the object can be put. Without the lens, the eye cannot focus on anything closer than about $25\ \text{cm}$, so that is the largest angle the object can be made to subtend. With the lens, the object can be brought much closer, to about the focal length, and still be seen sharply, because the lens places its image at a comfortable distance. The closer object subtends a larger angle than it would at $25\ \text{cm}$, and that ratio is the angular magnification.

Question 9.25 (b)

Solution. Yes, it decreases a little. With the eye at the lens, the image subtends at the eye exactly the angle the object subtends at the lens. With the eye moved back, both the image and the object are further from the eye, and the angle at the eye is slightly less than the angle at the lens. The effect is negligible when the image is very far away.

Question 9.25 (c)

Solution. Two things. Lenses of very short focal length are highly curved and hard to grind. More importantly, the aberrations of a lens, both spherical and chromatic, become much worse as its focal length shrinks, so the image blurs faster than it grows. In practice a single convex lens gives a magnifying power of about 3 at most; an aberration-corrected combination of lenses can raise this by a factor of about 10.

Question 9.25 (d)

Solution. The magnifying power is the product of the two lenses’ magnifications, and each is large only when its focal length is short. The eyepiece gives $m_e = 1 + 25/f_e$ (with $f_e$ in centimetres), which grows as $f_e$ falls. For the objective, the lens equation gives $v_o = f_o|u_o|/(|u_o| – f_o)$, so

$$m_o = \frac{v_o}{|u_o|} = \frac1{\left(|u_o|/f_o\right) – 1}$$

which is large when $|u_o|$ is only slightly greater than $f_o$. A microscope is used on objects very close to the objective, so $|u_o|$ is small, and $f_o$ has to be smaller still.

Question 9.25 (e)

Solution. Every ray that passes through the objective and then the eyepiece goes through the image of the objective formed by the eyepiece, a small bright disc called the eye-ring. It is the best place for the eye. With the pupil at the eye-ring and at least as large as it, the eye collects all the light the objective gathers and sees the whole field. With the eye pressed against the eyepiece instead, much of the light misses the pupil and the field of view shrinks.

The position of the eye-ring depends on the separation $L$ of the lenses: treating the objective as an object at distance $L$ from the eyepiece, the eye-ring is at $v = f_eL/(L – f_e)$ behind the eyepiece, only a little more than $f_e$ when the tube is long. In a real microscope this distance is built into the design of the instrument.

Question 9.26

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

Solution. Take the microscope in normal use, with the final image at $25\ \text{cm}$. Then the eyepiece’s share of the magnification is fixed:

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

so the objective must supply $m_o = 30/6 = 5$. The objective forms a real, inverted image, so $v_o = -5u_o$, and its lens equation gives

$$-\frac1{5u_o} – \frac1{u_o} = \frac1{1.25} \quad\Rightarrow\quad -\frac6{5u_o} = 0.8 \quad\Rightarrow\quad u_o = -1.5\ \text{cm}, \qquad v_o = 7.5\ \text{cm}$$

For the eyepiece, with $v_e = -25\ \text{cm}$:

$$\frac1{u_e} = -\frac1{25} – \frac15 = -\frac6{25}, \qquad |u_e| = \frac{25}6 = 4.17\ \text{cm}$$

The lenses must be $v_o + |u_e| = 7.5 + 4.17 = 11.67\ \text{cm}$ apart.

Note on the book’s answer. The key’s working line prints $\dfrac1{5u_O} – \dfrac1{u_O} = \dfrac1{1.25}$, which would give $u_O = -1.0\ \text{cm}$. Because the objective’s image is inverted, $v_O = -5u_O$ and the first term should carry a minus sign; the key’s values $u_O = -1.5\ \text{cm}$ and $v_O = 7.5\ \text{cm}$ are the correct ones.

Question 9.27

A small telescope has an objective lens of focal length $140\ \text{cm}$ and an eyepiece of focal length $5.0\ \text{cm}$. 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 ($25\ \text{cm}$)?

Question 9.27 (a)

Solution.

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

Question 9.27 (b)

Solution. The objective’s image of height $h’$ still forms at its focus, where it subtends $\alpha = h’/f_o$ at the objective, the same angle as the distant object. The eyepiece is now moved in so that this image sits a distance $|u_e|$ from it that puts the final image at $25\ \text{cm}$: $\dfrac1{|u_e|} = \dfrac1{f_e} + \dfrac1{25}$. The final image subtends $\beta = h’/|u_e|$ at the eye, so

$$m = \frac\beta\alpha = \frac{f_o}{|u_e|} = \frac{f_o}{f_e}\left(1 + \frac{f_e}{25}\right) = 28\times\left(1 + \frac{5}{25}\right) = 33.6$$

Note on the book’s answer. The key’s formula has $f_O/25$ inside the bracket, which would give $28\times(1 + 140/25) = 184.8$. The bracket should read $1 + f_e/25$; the key’s value, $33.6$, is correct.

Question 9.28

(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 $100\ \text{m}$ tall tower $3\ \text{km}$ 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 $25\ \text{cm}$?

Question 9.28 (a)

Solution. In normal adjustment the objective’s focus and the eyepiece’s focus coincide, so the separation is $f_o + f_e = 140 + 5.0$.

Question 9.28 (b)

Solution. The tower subtends $\alpha = 100/3000 = 1/30\ \text{rad}$ at the objective. At $3\ \text{km}$ the tower is effectively at infinity, so its image forms at the focal plane, $140\ \text{cm}$ from the lens, and subtends the same angle there:

$$\frac h{140} = \frac1{30}, \qquad h = 4.7\ \text{cm}$$

Question 9.28 (c)

Solution. The eyepiece now acts as a magnifying glass on that $4.7\ \text{cm}$ image, with the final image at $25\ \text{cm}$. As in 9.27(b), $\dfrac1{|u_e|} = \dfrac15 + \dfrac1{25} = \dfrac6{25}$, so the eyepiece’s linear magnification is

$$m_e = \frac{|v_e|}{|u_e|} = 25\times\frac6{25} = 6, \qquad h_{\text{final}} = 6\times\frac{140}{30} = 28\ \text{cm}$$

Question 9.29

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

Solution. Follow the light mirror by mirror, as with two lenses. The large concave mirror has $f_1 = 220/2 = 110\ \text{mm}$, so parallel light reflected from it would come to a focus $110\ \text{mm}$ in front of it. The small mirror sits only $20\ \text{mm}$ in front of the large one and intercepts the converging light first. That light is heading for a point $110 – 20 = 90\ \text{mm}$ behind the small mirror, which is therefore a virtual object for it.

For the small mirror, take the direction of the light arriving at it (away from the large mirror) as positive. The virtual object lies in that direction, $u = +90\ \text{mm}$, and a convex mirror has its focus behind it, $f_2 = +140/2 = +70\ \text{mm}$. The mirror equation gives

$$\frac1v = \frac1{f_2} – \frac1u = \frac1{70} – \frac1{90} = \frac{9 – 7}{630} = \frac2{630}, \qquad v = +315\ \text{mm}$$

The final image is $315\ \text{mm}$ from the small mirror. Its sign is positive, which places it behind the small mirror, on the side away from the large mirror.

Note on the book’s answer. The key gives the distance, $315\ \text{mm}$ from the smaller mirror, but not the side. The positive $v$ matters: a convex mirror of focal length $70\ \text{mm}$ diverges light more strongly than a beam converging to a point $90\ \text{mm}$ away can make up for, so the reflected light spreads out and only appears to come from a point $315\ \text{mm}$ behind the small mirror. With these numbers the image is virtual, not the real image behind the objective drawn in Fig. 9.26; that needs a secondary whose focal length exceeds its distance from the primary focus. It is sometimes placed $315\ \text{mm}$ from the small mirror on the eyepiece side, which the sign of $v$ does not allow.

Question 9.30

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 $1.5\ \text{m}$ away?

Solution. When a mirror turns through an angle $\theta$, its normal turns through $\theta$ too. A ray that was hitting the mirror along the normal now meets it at angle of incidence $\theta$ and leaves at angle $\theta$ on the other side of the normal, so the reflected ray has turned through $2\theta = 7^\circ$ from its old direction. On a screen $1.5\ \text{m}$ away the spot moves by

$$d = 1.5\tan7^\circ = 1.5\times0.1228 = 0.184\ \text{m}$$

This doubling is why a mirror and light beam make such a sensitive pointer for a galvanometer.

Question 9.31

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 $45.0\ \text{cm}$. The liquid is removed and the experiment is repeated. The new distance is measured to be $30.0\ \text{cm}$. What is the refractive index of the liquid?

Solution. The image coincides with the needle when the light leaving the lens system strikes the mirror normally: then it retraces its path exactly and comes back to a focus where it started. Light leaves the system parallel to the axis only if it started from the focus, so each measured distance is the focal length of whatever lies between the needle and the mirror.

Without the liquid only the glass lens has power (the thin film of air does nothing), so $f_1 = 30.0\ \text{cm}$. The lens maker’s formula for an equiconvex lens then gives the radius of its faces:

$$\frac1{30} = (1.50 – 1)\frac2R \quad\Rightarrow\quad R = 30\ \text{cm}$$

With the liquid the system is the glass lens in contact with a lens of liquid, and the combined focal length is $F = 45.0\ \text{cm}$. Powers of lenses in contact add, so the liquid lens has

$$\frac1{f_2} = \frac1F – \frac1{f_1} = \frac1{45} – \frac1{30} = -\frac1{90}, \qquad f_2 = -90\ \text{cm}$$

The liquid fills the space between the lower face of the lens and the flat mirror, so it is a plano-concave lens: one face has radius $30\ \text{cm}$, curving towards the incoming light ($R_1 = -30\ \text{cm}$), and the other is flat ($R_2 = \infty$). Its lens maker’s formula gives

$$-\frac1{90} = (n – 1)\left(-\frac1{30} – 0\right) \quad\Rightarrow\quad n – 1 = \frac{30}{90} = \frac13, \qquad n = 1.33$$

The value suggests the liquid is water.

Common mistakes

• Question 9.1: giving the concave mirror a positive focal length. The focus of a concave mirror is in front of it, against the incident light, so $f = -18\ \text{cm}$. With $f = +18\ \text{cm}$ the equation gives a virtual image $10.8\ \text{cm}$ behind the mirror, which no screen could catch. That result contradicts the question, and should prompt a check of the signs.
• Question 9.3: moving the microscope the wrong way. In the denser liquid the needle’s image rises towards the surface, closer to the microscope, so the microscope has to be raised by $1.7\ \text{cm}$, not lowered. Students often compute the new apparent depth correctly and then forget what it means for the instrument.
• Question 9.4: multiplying the two refractive indices. The index of glass relative to water is $n_{ga}/n_{wa}$, a ratio, because both are measured against air and the air cancels. Multiplying them gives an index of about $2.0$ and $r \approx 21^\circ$, far too much bending for a surface between two media whose indices are as close as water’s and glass’s.
• Question 9.8: treating $P$ as a real object. The light has not come from $P$; it is heading towards it, so $u = +12\ \text{cm}$. Using $u = -12\ \text{cm}$ turns (a) into a virtual image $30\ \text{cm}$ in front of the lens, an answer that describes a different experiment.
• Question 9.11: swapping the two eyepiece formulae. The eyepiece gives $1 + D/f_e$ when the final image is at $25\ \text{cm}$ and $D/f_e$ when it is at infinity. Mixing them up changes the answers to $16$ and $16.9$, and the object distance must also be recalculated in (b), because the intermediate image moves to the eyepiece’s focus.
• Question 9.17: using the angle at the end face as the angle at the wall. The ray meets the end face and the side wall at different angles: inside the fibre $i’ = 90^\circ – r$. Setting $r$ itself equal to the critical angle gives the wrong limit.
• Question 9.22: treating magnification and magnifying power as the same thing. The image is ten times the size of the card but only $2.8$ times bigger in apparent size, because it is also ten times further away. What the eye perceives is the angle, not the size.
• Question 9.30: using $3.5^\circ$ instead of $7^\circ$. When the mirror turns through $\theta$, the reflected ray turns through $2\theta$. Using the mirror’s own deflection halves the answer, to $9.2\ \text{cm}$.

Practise next

• Chapter 10, Wave Optics — interference, diffraction and polarisation, the effects that ray optics cannot explain and that set the limit on a telescope’s resolution.
• Chapter 8, Electromagnetic Waves — light as an electromagnetic wave, and the link between refractive index and the speed of light in a medium that underlies Snell’s law.