Class 12 Physics Chapter 9 NCERT Solutions: Ray Optics and Optical Instruments Questions and Formulas

9.1 Size of the candle, h = 2.5 cm

Let the image size be = h’

Object distance, u = $-$ 27 cm

Radius of curvature of the concave mirror, R = $-$ 36 cm

Focal length of the concave mirror, f = $\frac{R}{2}$ = $-$ 18 cm

Image distance = v

The image distance can be obtained by using mirror formula: $\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

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

v = −54 cm

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as:

m = $\frac{h\text{'}}{h}$ = $-\frac{v}{u}$

h’ = $-\frac{v}{u}$ $\times h$ = $-$ $\frac{-54}{-27}$ $\times 2.5$ = - 5 cm

The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual. If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

9.2 This is an NCERT question from the class 12th Physics chapter 9 called 'Ray Optics and Optical instruments'. We will be explaining this question step-by-step for the convinience of students:

• Height of the needle, $h_1$ = 4.5 cm
• Object distance, u = - 12 cm
• Focal length of the convex mirror, f = 15 cm
• Image distance = v

The value of v can be obtained using the mirror formula

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

v = $\frac{60}{9}$ = 6.7 cm

Hence, the image of the needle is 6.7 cm away from the mirror. Also it is on the other side of the mirror.

The image size ( $h_2)$ is given by the magnification formula (m) as:

m = $\frac{h_2}{h_1}$ = $-\frac{v}{u}$ = $-\frac{6.7}{-12}$ = $\frac{6.7}{12}$

$h_2=\frac{6.7}{12}\times 4.5$ = 2.5 cm

Hence m = $\frac{2.5}{4.5}$ = 0.56

The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual and diminished. If the needle is moved farther from the mirror, the image will also move away from the mirror and the size of the image will reduce gradually.

While this is one of the questions from the chapter, we have an NCERT excercise on Ray optics and optical instruments. Students who are currently in class 12th and plan on to take entrance exams, must go through this excercise for understanding how they can solve each question in proper steps.

9.3 Actual depth of the needle in water, $h_1$ = 12.5 cm

Apparent depth of the needle in water, $h_2$ = 9.4 cm

Refractive index of water =, it can be calculated as $\mu$ = $\frac{h_1}{h_2}$

So $\mu$ = $\frac{12.5}{9.4}$ = 1.33

So the refractive index of water = 1.33

The water is replaced by a liquid with refractive index, $\mu \text{'}$ = 1.63

From the relation of, ${\mu }^{\text{'}}$ = $\frac{h_1}{h_2^{\text{'}}}$ , where $h_2^{\text{'}}$ is the new apparent depth by microscope, we get

$h_2^{\text{'}}=$ $\frac{h_1}{{\mu }^{\text{'}}}$ = $\frac{12.5}{1.63}$ = 7.67 cm

So to focus again, microscope needs to be moved up by 9.4 – 7.67 cm = 1.73 cm

9.4 In figure (a) – Glass-Air interface:

Angle of incidence, i = 60 $°$ , Angle of refraction, r = 35 $°$

The relative refractive index of glass with respect to air is given by Snell’s law as:

${\mu }_g^a$ = $\frac{\sin ?i}{\sin ?r}$ = $\frac{\sin ?60°}{\sin ?35°}$ = 1.51 ………(1)

In figure (b) – Air - Water interface:

Angle of incidence, i = 60 $°$ , Angle of refraction, r = 47 $°$

The relative refractive index of water with respect to air is given by Snell’s law as:

${\mu }_w^g$ = $\frac{\sin ?i}{\sin ?r}$ = $\frac{\sin ?60°}{\sin ?47°}$ = 1.18 ………(2)

Using equation (1) and (2), the relative refractive index of glass with respect to water can be obtained as

${\mu }_g^w$ = $\frac{{\mu }_g^a}{{\mu }_w^g}$ = $\frac{1.51}{1.18}$ = 1.28

In figure (c) – Water - Glass interface:

Angle of incidence, i = 45 $°$ , Angle of refraction = r

The relative refractive index of glass with respect to water is given by Snell’s law as:

${\mu }_g^w$ = $\frac{\sin ?i}{\sin ?r}$ = $\frac{\sin ?45°}{\sin ?\mathrm{r}}$ = 1.28

sin r = $\frac{\sin ?45°}{1.28}$

r = 33.53^$°$

Hence, the angle of refraction at the water – glass interface is 33.53^$°$

9.5 Actual depth of the bulb in water, $d_1$ = 80 cm = 0.8 m

Refractive index of water, $\mu$ = 1.33

In the given figure, i = angle of incidence, r = angle of refraction = 90 $°$

The light source, bulb (B) is placed at the bottom of the tank.

Since the bulb is a point source, the emergent light can be considered as a circle of radius, with radius R = $\frac{AC}{2}$ = OA = OB

Using Snell’s law, we can write the refractive index of water as:

$\mu =\frac{\sin ?r}{\sin ?i}$ = $\frac{\sin ?90°}{\sin ?i}$ = 1.33

i = 48.75 $°$

In Δ OBC, $\tan ?i$ = $\frac{OC}{OB}$

$\tan ?48.75°$ = $\frac{R}{80}$

R = 91.23 cm

Hence, the area of the water surface = $\pi R^2$ = 2.61 $\times {10}^4$ ${cm}^2$ = 2.61 $m^2$

9.6 The angle of minimum deviation, ${\delta }_m$ = 40 $°$

Angle of prism, A = 60 $°$

Let the refractive index of water, $=1.33$ , and the refractive index of prism material = $\mu \text{'}$

The angle of deviation is related to refractive index $\mu \text{'}$ is given as

${\mu }^{\text{'}}=\frac{\sin ?\frac{(A+{\delta }_m)}{2}}{\sin ?\frac{A}{2}}$ = $\frac{\sin ?\frac{(60°+40°)}{2}}{\sin ?\frac{60°}{2}}$ = $\frac{\sin ?50°}{\sin ?30°}$ = 1.532

So the refractive index of prism material is 1.532

Since the prism is placed in water, let ${\delta }_m^{\text{'}}$ be the new angle of minimum deviation.

The refractive index of glass with respect to water is given by the relation:

${\mu }_g^w$ = $\frac{{\mu }^{\text{'}}}{\mu }$ = $\frac{\sin ?\frac{(A+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{A}{2}}$

$\frac{1.532}{1.33}$ = $\frac{\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{60°}{2}}$

1.152 $\times \sin ?30°$ = $\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}$

$\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}={\sin }^{-1}?0.576$ = 35.2 $°$

${\delta }_m^{\text{'}}=2\times$ 35.2 $°$ - 60 $°$ = 10.33 $°$

Hence the new minimum angle of deviation is 10.33$°$

9.7 Refractive index of glass, $\mu$ = 1.55

Focal length required for the double-convex lenses, f = 20 cm

Let the radius of curvature of one face of the lens be = $R_1$ and the other face be = $R_2$

Let the radius of curvature of the double convex lenses be = R

Then $R_1=R$ and $R_2=-R$

The value of R can be calculated as:

$\frac{1}{f}$ = ( $\mu$ - 1) $\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

$\frac{1}{20}$ = ( $1.55$ - 1) $\left[\frac{1}{R}+\frac{1}{R}\right]$

0.05 = 0.55 $\times \frac{2}{R}$

R = 22 cm

Hence, the radius of curvature for double-convex lens is 22 cm.

9.8 Object distance, u = 12 cm

Focal length of the convex lens, f = 20 cm

Image distance = v

According to lens formula

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ - $\frac{1}{12}$ = $\frac{1}{20}$

$\frac{1}{v}$ = $\frac{1}{20}+\frac{1}{12}$ or v = $\frac{60}{8}$ = 7.5 cm

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{}7.5\mathrm{}\mathrm{c}\mathrm{m}\mathrm{}\mathrm{a}\mathrm{w}\mathrm{a}\mathrm{y}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{s},\mathrm{}\mathrm{t}\mathrm{o}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{t}\mathrm{s}\mathrm{}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}.$

Focal length of the concave lens, f = -16 cm

Image distance = v

According to lens formula

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ - $\frac{1}{12}$ = $-\frac{1}{16}$

$\frac{1}{v}$ = $-\frac{1}{16}+\frac{1}{12}$ or v = $\frac{48}{1}$ = 48 cm

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{}48\mathrm{}\mathrm{c}\mathrm{m}\mathrm{}\mathrm{a}\mathrm{w}\mathrm{a}\mathrm{y}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{s},\mathrm{}\mathrm{t}\mathrm{o}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{t}\mathrm{s}\mathrm{}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}.$

9.11 Focal length of the objective lens, $f_1$ = 2.0 cm

Focal length of the eyepiece, $f_2$ = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

Least distance of distinct vision, d’ = 25 cm

Hence, image distance for the eyepiece, $v_2$ = - 25 cm

Let the object distance for the eyepiece be = $u_2$

According to lens formula, we get

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{-25}-\frac{1}{6.25}$

$u_2=-5cm$

Image distance for the objective lens, $v_1$ = d + $u_2$ = 15 – 5 = 10 cm

Let the object distance for the eyepiece be = $u_1$

According to lens formula, we get

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{10}-\frac{1}{2}$

$u_1=-2.5cm$

Magnitude for the object distance, $\left|u_1\right|$ = 2.5 cm

The magnifying power of a compound microscope is given by the relation,

m = $\frac{v_1}{\left|u_1\right|}$ ( 1 + $\frac{d\text{'}}{f_2\mathrm{}}$ ) = $\frac{10}{2.5}$ ( 1 + $\frac{25}{6.25}$ ) = 20

Hence the magnifying power of the microscope is 20

The final image is formed at infinity.

Hence, $v_2$ = $\infty$

Let the object distance for the eyepiece be = $u_2$

According to lens formula, we get

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{\infty }-\frac{1}{6.25}$

$u_2=-6.25cm$

Image distance for the objective lens, $v_1$ = d + $u_2$ = 15 – 6.25 = 8.75 cm

Let the object distance for the eyepiece be = $u_1$

According to lens formula, we get

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{8.75}-\frac{1}{2}$

$u_1=-2.59cm$

Magnitude for the object distance, $\left|u_1\right|$ = 2.59 cm

The magnifying power of a compound microscope is given by the relation,

m = $\frac{v_1}{\left|u_1\right|}\times$ $\frac{d\text{'}}{\left|u_2\right|}$ = $\frac{8.75}{2.59}\times \frac{25}{6.25}$ = 13.51

Hence the magnifying power of the microscope is 13.51.

9.10 Focal length of the convex lens,  $f_1$ = 30 cm, focal length of the concave lens,  $f_2$ = -20 cm

Let the focal length of the combined lens be = f

The equivalent focal length of a system of two lenses in combined form is given by

$\frac{1}{f}$ = $\frac{1}{f_1\mathrm{}}$ + $\frac{1}{f_2}$

$\frac{1}{f}$ = $\frac{1}{30}$ - $\frac{1}{20}$ = $-\frac{1}{60}$

f = $-$ 60 cm

Hence, the focal length of the combination lenses is 60 cm. The negative sign of lenses acts as a diverging lens.

9.12 Focal length of the objective lens, $f_o$ = 8 mm = 0.8 cm

Focal length of the eyepiece, $f_e$ = 2.5 cm

Object distance for the objective lens, $u_o$ = - 9.0 mm = -0.9 cm

Least distance of distant vision, d = 25 cm

Image distance of the eyepiece, $v_e$ = -d = -25 cm

Object distance of the eyepiece = $u_e$

Using the lens formula, we can obtain the value of $u_e$ as:

$\frac{1}{v_e}$ - $\frac{1}{u_e}$ = $\frac{1}{f_e}$ or $\frac{1}{u_e}$ = $\frac{1}{v_e}-\frac{1}{f_e}$ = $\frac{1}{-25}-\frac{1}{2.5}$

$u_e=-2.27cm$

Using the lens formula, we can obtain the value of $v_o$ as:

$\frac{1}{v_o}$ - $\frac{1}{u_o}$ = $\frac{1}{f_o}$ or $\frac{1}{v_o}$ = $\frac{1}{u_o}+\frac{1}{f_o}$ = $\frac{1}{-0.9}+\frac{1}{0.8}$

$v_o=7.2cm$

The distance between the objective lens and the eyepiece = $\left|u_e\right|$ + $v_o$ = 2.27 + 7.2 = 9.47 cm

The magnifying power of microscope is calculated as:

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

Hence the magnification of the microscope is 88.

9.13 Focal length of the objective lens,  $f_o$ = 144 cm

Focal length of the eyepiece,  $f_e$ = 6.0 cm

The magnifying power of the telescope, m = $\frac{f_o}{f_e}$ = $\frac{144}{6}$ = 24

The separation between the eyepiece and objective lens = $f_e+f_o$ = 6 + 144 = 150 cm

9.14 Focal length of the objective lens, $f_o$ = 15 m = 1500 cm

Focal length of the eyepiece, $f_e$ = 1.0 cm

The angular magnification of the telescope is given by,

m = $\frac{f_o}{f_e}$ = $\frac{1500}{1}$ = 1500

Hence the angular magnification of the telescope is 1500

Diameter of the moon, $d_m$ = 3.48 $\times {10}^6$ m

Radius of the lunar orbit, $r_l$ = 3.8 $\times {10}^8$ m

Let $d_1$ be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image. Hence,

$\frac{d_m}{r_l}$ = $\frac{d_1\mathrm{}}{f_o\mathrm{}}$

$d_1=\frac{d_m\times f_o}{\mathrm{}{r}_l}$ = $\frac{3.48\mathrm{}\times {10}^6\times 1500\mathrm{}}{3.8\mathrm{}\times {10}^8}$ = 13.74 cm

Hence the diameter of the moon’s image is 13.74 cm.

9.15 For a concave mirror, the focal length, f < 0

When the object is placed on the left side of the mirror, the object distance, u is negative

For image distance v, we can write the lens formula as:

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$ (since u is negative) ……….(1)

The object lies between f and 2f, i.e. 2f < u < f

or, $\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}$

or, $-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}$

or, $\frac{1}{f}$ $-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}$ < 0 …………(2)

Using equation (1), we get

$\frac{1}{2f}<\frac{1}{v}$ < 0

Therefore, $\frac{1}{v}$ is negative, hence v is negative

$\frac{1}{2f}<\frac{1}{v}or$ 2f > v or –v > -2f . Hence, the image lies beyond 2f

For a convex mirror, the focal length, the focal length f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative (< 0)

For image distance v, we have the mirror formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

Using equation (2), we can conclude that:

$\frac{1}{v}<0$ or v > 0

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

For a convex mirror, the focal length, the focal length f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative (< 0)

For image distance v, we have the mirror formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

But we have u < 0, then

$\frac{1}{v}>\frac{1}{f}$ or v < f

$\mathrm{H}$ ence, the image formed is diminished and it is located between the focus (f) and the pole.

For a concave mirror, the focal length (f) is negative, f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative, u < 0

It is placed between the focus (f) and the pole, so f > u > 0

So, $\frac{1}{f}$ < $\frac{1}{u}$ < 0 or $\frac{1}{f}$ - $\frac{1}{u}$ < 0

For image distance v, we have the formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}$ < 0 or v > 0

The image is formed on the right side of the mirror, hence it is a virtual image.

For u < 0 and v > 0, we can write $\frac{1}{u}$ > $\frac{1}{v}$ or v > u

Magnification, m = $\frac{v}{u}$ > 1

Hence, the formed image is enlarged.

9.16 The actual depth of the pin, d = 15 cm

Let the apparent depth of the pin be = d’

Refractive index of the glass, $\mu$ = 1.5

Ratio of actual depth to the apparent depth is equal to the refractive index of the glass. i.e.

$\mu =\frac{d}{d\text{'}}$ or

d’ = $\frac{d}{\mu }$ = $\frac{15}{1.5}$ = 10

The distance at which the pin appears to be raised = d – d’ = 15 – 10 = 5 cm

For a small angle of incidence, this distance does not depend upon the location of the slab.

9.17 Refractive index of the glass fibre, ${\mu }_1$ = 1.68

Refractive index of outer covering of the pipe, ${\mu }_2$ = 1.44

Given:-

Angle of incidence = i, Angle of refraction = r, Angle of incidence at the interface = i’

The refractive index of the inner core-outer core interface, $\mu$ is given as

$\mu =\frac{{\mu }_2}{{\mu }_1}$ = $\frac{1}{\sin ?i\text{'}}$

$\frac{1}{\sin ?i\text{'}}=\frac{1.44}{1.68}$ or $\sin ?i\text{'}$ = $\frac{1.44}{1.68}$

i’ = 59 $°$

For the critical angle, total reflection take place only when i > i’ i.e. i > 59 $°$

Maximum angle of reflection, ${\tau }_{max}$ = 90 $°-i\text{'}$ = 31 $°$

Let $i_{max}$ be the maximum angle of incidence and $r_{max}$ be the maximum angle of reflection.

${\mu }_1=\frac{\sin ?i_{max}}{\sin ?r_{max}}$ or $\sin ?i_{max}={\mu }_1\times \sin ?r_{max}$

$\sin ?i_{max}=$ 1.68 $\times sin31°$

$i_{max}=59.91°$ $\approx 60°$

The entire rays incident at angles lying in the range of 0 < i < 60 will suffer total internal reflection.

If the outer covering is absent, then:

Refractive index of the outer pipe, ${\mu }_1$ = Refractive index of air = 1

For the angle of incidence, i = 90 $°$ , we can write Snell’s law at ‘air – pipe’ interface as :

$\frac{sini}{\sin ?r}$ = ${\mu }_2$ = 1.68

$\sin ?r=$ $\frac{\sin ?i}{1.68}$ = $\frac{\sin ?90°}{1.68}$ = $\frac{1}{1.68}$

r = 36.53 $°$

i’ = 90 $°-$ 36.53 $°$ = 53.47 $°$

Since i’ > r, all incident rays will suffer total internal reflection.

9.18 (a) Yes, Plane and convex mirrors can produce real images as well, if the object is virtual.

i.e. if the light rays converging at a point behind a plane mirror or a convex mirror, they are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.

(b) No. A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.

(c) Water being the denser medium than air, the light rays will deviate and the fisherman will appear to be taller to the diver.

(d) The apparent depth of the tank water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near normal viewing.

(e) No, diamond is used as cutter because of its hardness, not refractive index.

9.19 Distance between the object and the image, d = 3 m

Let maximum focal length be $f_{max}$

For real image, the maximum focal length is given as:

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

Hence, for this purpose, the maximum possible focal length is 0.75 m

9.20 Distance between the object and the image ( screen), D = 90 cm

Distance between two locations of the convex lens, d = 20 cm

Let the focal length of the lens be = f

Focal length is related to D and d as:

f = $\frac{D^2-d^2}{4D}$ = $\frac{{90}^2-{20}^2}{4*90}$ = 21.39 cm

Therefore, the focal length of the convex lens is 21.39 cm

9.21 Focal length of the convex lens, $f_1$ = 30 cm

Focal length of the concave lens, $f_2$ = -20 cm

Distance between two lenses, d = 8.0 cm

When the parallel beam of light is incident on the convex lens first:

According to lens formula, we have:

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ , where u = object distance = $\infty$ and $v_1$ = Image distance

$\frac{1}{v_1}=$ $\frac{1}{f_1}+\frac{1}{u_1}$ = $\frac{1}{30}$ + $\frac{1}{\infty }$

$v_1=30cm$

The image will act as a virtual object for the concave lens. Applying lens formula to the concave lens, we have:

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ ,

where $u_2$ = object distance = $v_1-d$ = 30 – 8 = 22 cm and

$v_2$ = image distance

$\frac{1}{v_2}=\frac{1}{f_2}+$ $\frac{1}{u_2}$ = $\frac{1}{-20}$ + $\frac{1}{22}$

$v_2=$ - 220 cm

The parallel incident beam appears to diverge from a point that is 220 - $\frac{d}{2}$ = 216 cm from the centre of the combination of two lenses.

When the parallel beam of light is incident, from the left on the concave lens first:

According to the lens formula,

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$

$\frac{1}{v_2}=\frac{1}{f_2}+\frac{1}{u_2}$ , where $u_2$ = Object distance = - $\infty$

$\frac{1}{v_2}=\frac{1}{*20}+\frac{1}{-\mathrm{}\infty }$

$v_2=$ - 20 cm

The image will act as a real object for the convex lens.

Applying lens formula to the convex lens, we have

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ , where $u_1$ = object distance = - (20 + d) = - 28 cm and $v_1$ = Image distance

$\frac{1}{v_1}=\frac{1}{f_1}$ + $\frac{1}{u_1}$ = $\frac{1}{30}$ + $\frac{1}{-28}$

$v_1$ = - 420 cm

Hence, the parallel incident beam appear to diverge from a point that is 420 – 4 = 416 cm from the left of the centre of the combination of the two lenses.

The answer does not depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.

Height of the image, $h_1$ = 1.5 cm

Object distance from the side of the convex lens, $u_1$ = - 40 cm

$\left|u_1\right|$ = 40 cm

According to lens formula:

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ , where $v_1$ = Image distance

$\frac{1}{v_1}$ = $\frac{1}{f_1}+\frac{1}{u_1}$ , where $v_1$ is the image distance

$\frac{1}{v_1}$ = $\frac{1}{30}+\frac{1}{-40}$

$v_1$ = 120 cm

$\mathrm{M}\mathrm{a}\mathrm{g}\mathrm{n}\mathrm{i}\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n},m_1=$ $\frac{v_1}{\left|u_1\right|}$ = 3

Hence the magnification due to the convex lens is 3.

The image formed by the convex lens acts as an object for the concave lens.

According to lens formula,

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ , where $u_2$ is the object distance = + (120 – 8) = + 112 cm

$\frac{1}{v_2}=$ $\frac{1}{f_2}+\frac{1}{u_2}$ = $\frac{1}{-20}$ + $\frac{1}{112}$

$v_2$ = - 24.35 cm

Magnification, $m_2$ = $\left|\frac{v_2}{112}\right|$ = 0.217

The magnification of combination of two lenses given as m = $m_1\times m_2$ = 0.652

The magnification of the combination is also expressed as

$\frac{h_2}{h_1}$ = 0.652, $h_2$ = 0.652 $\times h_1$ = 0.652 $\times 1.5$ = 0.978 cm

Hence, the height of the image is 0.978 cm.

9.22 The incident, refracted and emergent rays associated with a glass prism ABC is shown in the adjoined figure.

Angle of the prism, $\angle A$ = 60 $°$

Refractive index of the prism, $\mu$ = 1.524

Let $\angle i_1$ be the incident angle, ${\angle r}_1$ be the refracted angle, $\angle r_2$ be the incidence angle on face AC and $\angle e$ be the emergent angle from the prism = 90 $°$

According to Snell’s law, for face AC, we can have:

$\frac{\sin ?e}{\sin ?r_2}$ = $\mu$

$\sin ?r_2=\frac{\sin ?e}{\mu }$ = $\frac{\sin ?90°}{1.524}$ = 0.656

${\angle r}_2$ = 41 $°$

From the Δ ABC, $\angle A=$ $\angle r_1$ + ${\angle r}_2$ . Hence ${\angle r}_1$ = 60 $°-$ 41 $°=19°$

According to Snell’s law, for face AB, we have

$\mu =\frac{sin{i}_1}{\sin ?r_1}$ or $\sin ?i_1=\mu \times \sin ?r_1$ = 1.524 $\times \sin ?19°=0.496$

$\angle i_1$ = 29.75 $°$

Hence the angle of incidence is 29.75^$°$

9.23 Area of each square, A = 1 ${mm}^2$

Object distance, u = - 9 cm

Focal length of the converging lens, f = 10 cm

For image distance v, the lens formula can be written as

$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

$\frac{1}{10}$ = $\frac{1}{v}$ - $\frac{1}{-9}$

$\frac{1}{v}$ = $\frac{1}{10}-\frac{1}{9}$

v = -90 cm

Magnification, m = $\frac{v}{u}$ = $\frac{-90}{-9}$ = 10

Therefore the area of each square of the virtual image

= 10 $\times 10{mm}^2$ = 100 ${mm}^2=1{cm}^2$

Magnifying power of the lens = $\frac{d}{\left|u\right|}$ = $\frac{25}{9}$ = 2.8

The magnification in (a) is not the same as the magnifying power in (b). The magnification magnitude is = $\frac{v}{u}$ and the magnifying power is = $\frac{d}{\left|u\right|}$ . The two quantities will be equal when the image is formed at the near point (25 cm)

9.24 The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm)

Image distance, v = -d = -25 cm

Focal length, f = 10 cm

Object distance = u

According the lens formula, we have:

$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

$\frac{1}{u}=\frac{1}{v}-\frac{1}{f}$ = $\frac{1}{-25}$ - $\frac{1}{10}$ = - $\frac{7}{50}$

u = - 7.14 cm

Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.

Magnification= $\left|\frac{v}{u}\right|$ = $\frac{25}{7.14}=3.50$

Magnifying power = $\frac{d}{\left|u\right|}$ = $\frac{25}{7.14}$ = 3.50

9.25 Area of the virtual image of each square, A = 6.25 ${mm}^2$

Area of each square,  $A_0$ = 1 ${mm}^2$

Hence the linear magnification of the object can be calculated as:

m = $\sqrt{\frac{A}{A_0}}$ = $\sqrt{\frac{6.25}{1}}$ = 2.5, but m = $\frac{v}{u}$ or v = 2.5u

Focal length of the magnifying glass, f = 10 cm. According to lens formula

$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

$\frac{1}{10}$ = $\frac{1}{2.5u}$ - $\frac{1}{u}$ or $\frac{1}{10}=-\frac{1.5}{2.5u}$ . Hence u = -6 cm and v = -15 cm

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye. Hence, it cannot be seen by the eyes directly.

9.26 Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the object placed closer than the least distance of distinct vision (i.e. 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.

Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is lightly less than the angle subtended at the lens. Image distance does not have any effect on angle magnification.

The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal length is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.

The angular magnification produced by the eyepiece of a compound microscope is [( $\frac{25}{f_e}$ ) + 1], where $f_e$ = focal length of the eyepiece