NCERT Solution | Ray Optics and Optical Instruments

Q:1. 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? Described the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved ?

Sol. Given, radius of curvature of concave mirror, R = – 36 cm (for concave mirror, radius of curvature is taken as negative).

Focal length, f = R/2 = -36/2 = -18 cm , Height of object, h = 2.5 cm

Using the mirror formula,

$\large \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\large \frac{1}{v} + \frac{1}{-27} = \frac{1}{-18}$

v = -54 cm
Screen should be placed at 54 cm from the mirror on the same side as the object .

$\large m = \frac{h_I}{h_O} = -\frac{v}{u}$

$\large \frac{h_I}{2.5} = -\frac{-54}{-27}$

hI = -5 cm

The negative sign shows that the image is formed in front of the mirror and it is inverted. Thus, the screen should be placed at a distance 54 cm and the size and the nature of image 5 cm, real, inverted and magnified in nature.

If we move the object near to the mirror (as u→f,v→∞), the screen should be moved away from mirror, as, the distance of object is less than focal length, (u<f) no screen is required, because the image formed is virtual.

Q:2. 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.

Sol.
Given, focal length of convex mirror, f = +15 cm.

Distance of object, u = -12 cm

Size of object, h = 4.5 cm

Using the mirror formula,

$\large \frac{1}{v} + \frac{1}{-12} = \frac{1}{15}$

v = 60/9 = 6.7 cm

Image is virtual & formed at 6.7 cm behind the mirror.

$\large m = \frac{h_I}{h_O} = -\frac{v}{u}$

$\large m = \frac{h_I}{4.5} = -\frac{6.7}{-12}$

hI = 2.5 cm

As, hI is positive, so image is erect and virtual.

As, the needle moves away from the mirror, the image also moves away from the mirror (as u→∞,v →f) and the size of image goes on decreasing.

Q:3. 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 upto the same height, by what distance would the microscope have to be moved to focus on the needle again?

Sol : Case I : When tank is filled with the water.

Given, the apparent depth = 9.4 cm

So, real depth = 12.5 cm

Refractive index of water, $\large \mu_w = \frac{Real \; depth}{Apparent \; depth}$

= 12.5/9.4 =1.33

Case II : When tank is filled with the liquid.

Refractive index of liquid, μl =1.63

$\large \mu_l = \frac{Real \; depth}{Apparent \; depth}$

⇒ 1.63=12.5/(Apparent depth

Apparent depth=12.5/1.63 = 7.67 cm

⇒ The microscope is shifted by 9.4 -7.67 = 1.73 cm.

Q:5 . 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 ? R.I of water is 1.33 . Consider the bulb be a point source .

Sol: Applying Snell’s law for water-air interface

$\large \mu_w sini = 1 \times sin90^o$

Sini = 1/1.33 = 3/4

$\large tani = \frac{r}{h}$

r = h tani

Area of the surface of water , A =  πr2

Q: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 cm?

Sol: μ = 1.55 , R1 = R & R2 = -R

$\large \frac{1}{f} = (\mu -1) (\frac{1}{R_1} – \frac{1}{R_2})$

$\large \frac{1}{20} = (1.55 -1) (\frac{1}{R} + \frac{1}{R})$

R = 22 cm

Q:8. 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 (i) a convex lens of focal length 20 cm, and (ii) a concave lens of focal length 16 cm?

Sol.  Here, the point P on the right side of the lens acts as virtual object.

Given, distance of object from the lens, u=12 cm Focal length of convex lens,f=+ 20 cm

Using lens formula,

$\large \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

$\large \frac{1}{v} – \frac{1}{12} = \frac{1}{20}$

⇒ v=7.5 cm

Thus, the beam converges on the right side of lens at a distance of 7.5 cm.

Distance of object from the lens, u=12 cm

Using lens formula, $\large \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

$\large \frac{1}{v} – \frac{1}{12} = \frac{1}{-16}$

⇒ v = 48 cm

Thus, the beam converges on the right side of lens at a distance of 48 cm.

Q: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?

Sol: Size of object, O = 3 cm

Focal length of lens,f=-21 cm

Distance of object from the concave lens,u=-14 cm

Using lens formula,

$\large \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

$\large \frac{1}{v} + \frac{1}{14} = -\frac{1}{21}$

v = -8.4 cm

Using the formula of magnification.

$\large m = \frac{v}{u} = \frac{I}{O}$

Where, I=height of image

$\large \frac{-8.4}{-14} = \frac{I}{3}$

I = 1.8 cm

I is positive, therefore the image formed is virtual and erect, at distance of 8.4 cm form the lens on the same distance of object and height of image is 1.8 cm.

If the object is moved further away from the lens, the image moves towards the lens (never beyond focus). The size of image decrease gradually.

Q:10. What is the focal length of a convex of 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.

Sol: Given,focal length of convex lens, f1 =30 cm

Focal length of concave lens, f2 =-20 cm

Using the formula of combination of lenses,

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

$\large \frac{1}{f} = \frac{1}{30} – \frac{1}{20}$

f = -60 cm

Since, the focal length of combination is negative in nature. So, the combination behaves like a diverging lens, i.e. as a concave lens.

Q:11. 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 (i) the least distance of distinct vision (ii) at infinity ?
What is the magnifying power of the microscope in each case ?

Sol: fo = 2.0 cm , fe = 6.25 cm , uo = ?

(i) ve = -25 cm

$\large \frac{1}{v_e} – \frac{1}{u_e} = \frac{1}{f_e}$

$\large \frac{1}{-25} – \frac{1}{u_e} = \frac{1}{6.25}$

ue = – 5 cm

The distance between objective & eyepiece = 15 cm

vo = 15 – 5 = 10 cm

$\large \frac{1}{v_o} – \frac{1}{u_o} = \frac{1}{f_o}$

$\large \frac{1}{10} – \frac{1}{u_o} = \frac{1}{2}$

uo = -2.5 cm

Magnifying power , $\large M = \frac{v_o}{|u_o|}(1 + \frac{D}{f_e})$

$\large M = \frac{10}{2.5}(1 + \frac{25}{6.25})$

= 20

(ii) ve = ∞ – ue = fe = 6.25 cm

vo = 15 – 6.25 = 8.75 cm

$\large \frac{1}{v_o} – \frac{1}{u_o} = \frac{1}{f_o}$

$\large \frac{1}{8.75} – \frac{1}{u_o} = \frac{1}{2}$

uo = -17.5/6.75 = -2.59

Magnifying Power ,

$\large M = \frac{v_o}{|u_o|}(\frac{D}{u_e})$

$\large M = \frac{8.75}{2.59}(\frac{25}{6.25})$

= 13.51

q:12. A person with a normal near point (25 cm) using a compound microscope with an objective of focal length 8.0 mm and an eye piece of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus . What is the separation between the two lenses ? Calculate the magnifying power of the microscope ?

Sol: D = 25 cm , fo = 8.0 mm = 0.8 cm , fe = 2.5 cm

uo = -9.0 mm = 0.9 cm

$\large \frac{1}{v_e} – \frac{1}{u_e} = \frac{1}{f_e}$

$\large \frac{1}{-25} – \frac{1}{u_e} = \frac{1}{2.5}$

ue = -25/11 = – 2.27 cm

$\large \frac{1}{v_o} – \frac{1}{u_o} = \frac{1}{f_o}$

$\large \frac{1}{v_o} – \frac{1}{-0.9} = \frac{1}{0.8}$

vo = 0.72/0.1 = 7.2 cm

Separation between the two lenses = |ue| + vo

2.27 + 7.2 = 9.47 cm

Magnifying power , $\large M = \frac{v_o}{|u_o|}(1 + \frac{D}{f_e})$

$\large M = \frac{7.2}{0.9}(1 + \frac{25}{2.5})$

= 88

Q:13. A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6 cm. what is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Sol: Given, focal length of objective lens, fo =144 cm

Focal length of eyepiece, fe=6 cm

Magnifying power of the telescope in normal adjustment, (i.e. when the final image is formed at ∞)

m = – fo/fe =-144/6=-24

Separation between lenses,

L=fo + fe = 144+6=150 cm

Q:14 . (i)A giant reflecting telescope at an observatory an objective lens of focal length 15 m. if an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.

(ii)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 × 106 m and the radius of the lunar orbit is 3.8 × 108 m.

Sol: (i) Angular magnification $\large = \frac{f_o}{f_e} = \frac{15}{0.01}$

= 1500

(ii) If d is the diameter of the image , then

angle subtended by diameter of moon $\large = \frac{3.48 \times 10^6}{3.8 \times 10^8}$

angle subtended by image $\large = \frac{d}{15} = \frac{3.48 \times 10^6}{3.8 \times 10^8}$

d = 13.73 × 10-2 m = 13.73 cm

Q:15. Use the mirror equation to show that,
(i) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(ii) a convex mirror always produces a virtual image independent of the location of the object.
(iii) An object placed between the pole an focus of a concave mirror produces a virtual and enlarged image.

Sol: (i) The mirror formula

$\large \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

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

For concave mirror f < 0 , As object is on the left , u < 0

As object placed between f and 2f of a concave mirror

2f < u < f

$\large \frac{1}{2f}> \frac{1}{u} > \frac{1}{f}$

$\large – \frac{1}{2f} < -\frac{1}{u} < -\frac{1}{f}$

$\large \frac{1}{f} – \frac{1}{2f} < \frac{1}{f} -\frac{1}{u} < 0$

$\large \frac{1}{2f} < \frac{1}{v} < 0$

As v is negative & also v > 2 f i.e. image lies beyond 2f

(ii) For convex mirror, f > 0 , Also, u < 0

But from mirror equation,

$\large \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

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

$\large \frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

⇒ If f and u to be positive,

then 1/v > 0 ⇒ v > 0

Hence, virtual images is formed.

(iii) For concave mirror , f < 0

As object is placed between pole & focus

f < u < 0

$\large \frac{1}{f} – \frac{1}{u} > 0$

But $\large \frac{1}{f} – \frac{1}{u} = \frac{1}{v} > 0$

As v is positive , hence Image is on right .It must be virtual .

Also , $\large \frac{1}{v} < \frac{1}{|u|}$

i.e. v > |u|

Hence , Image is enlarged .

Q:16. A small pin fixed on a table top is viewed 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 is 1.5. Does the answer depend on the location of the slab ?

Sol: Real depth , x = 15 cm

Apparent depth , y = ?

$\large \mu = \frac{x}{y}$

$\large y = \frac{x}{\mu} = \frac{15}{1.5} = 10 cm$

Distance through which the dot appeared to be raised = x – y

=15 – 10 = 5 cm

Answer does not depend on the location of the slab .

Q:17. (i) Figure shows (see in Textbook) 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.

(ii) What is the answer, if there is no outer covering of the pipe ?

Sol: μ2 = 1.68 , μ1 = 1.44

$\large \frac{\mu_2}{\mu_1} = \frac{1}{sinC}$

$\large SinC = \frac{\mu_1}{\mu_2} = \frac{1.44}{1.68} = 0.8571$

C = 59°

Total internal reflection would takes place when i > C i.e. i > 59°

rmax = 90° – 59° = 31°

$\large \frac{sini_{max}}{sinr_{max}} = a_{\mu_g} = 1.68$

$\large sin i_{max} = 1.68 \times sin r_{max}$

$\large sin i_{max} = 1.68 \times sin31^o = 1.68 \times 0.5156$

= 0.8662

imax = 60°

Thus, all the rays which are incident in the range 0 < i < 60° , will suffer total internal reflection in the pipe (but i ≠ 0).

(ii) If there is no outer coating of pipe ,

μ2 = 1.68 , μ1 = 1

$\large sinC’ = \frac{1}{\mu} = \frac{\mu_1}{\mu_2}= \frac{1}{1.68}$

sinC’ = 0.5952

C’ = 36.5°

i = 90° will have r = 36.5°

i’ = 90° – 36.5° =53.5° , which is greater than C’ . Thus all rays incident at angles in the range 0 to 90° will suffer total internal reflection .

(i) you have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

(ii) A virtual image, we always say, cannot be caught on a screen. Yet when we see a virtual image, we are obviously bringing it on to the screen (i.e. the retina) of our eye. Is there a contradiction?

(iii) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman looks taller or shorter to the diver than what he actually is?

(iv) Does the apparent depth of a tank of water change, if viewed obliquely? If so, does the apparent depth increase or decrease?

(v) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?

Sol. (i) Yes, plane and convex mirrors can produce the real images, if the rays incident on the plane or convex mirror are converging to a point behind the mirror. A plane or convex mirror can produce a real image, if object is virtual.

(ii) No, there is no contradiction because virtual image formed by the spherical mirror acts as virtual object for eye lens. Our eye lens is convergent and it forms a real image of virtual object on retina.

(iii) As, the fisherman is in air, the rays of light travels from rarer to denser medium, so they bend toward the normal. Therefore, the fisherman appears taller.

(iv) Yes, the apparent depth decrease, further when water tank is viewed obliquely as compared to the depth when seen normally.

Q:19 . The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens . what is the maximum possible focal length of the lens required for the purpose ?

Sol: For a real image on the wall , minimum distance between the object and image should be 4f .

4f = 3

f = 3/4 = 0.75 m

Q:20 . 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.

Sol. Given, distance between screen and object, D = 90 cm

Using the displacement formula, $\large f = \frac{D^2 – d^2}{4 D}$

$\large f = \frac{90^2 – 20^2}{4 \times 90}$

f = 21.4 cm

Q: 21. (a) Determine the ‘effective focal length’ of the combination of two lenses of focal lengths 30 cm and – 20 cm if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side 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 convex lens in the above arrangement. The distance between the object and convex lens is 40 cm. Determine the magnification produced by the two lens system and size of image.

Sol: Here, f1 = 30 cm, f2 = 20 cm, d = 8.0 cm, f = ?
Late a parallel beam be incident on the convex lens first. If 2nd lens were absent, then
u1 = ∞ and f1 = 30 cm

$\displaystyle \frac{1}{v_1} – \frac{1}{u_1} = \frac{1}{f_1}$

$\displaystyle \frac{1}{v_1} – \frac{1}{\infty} = \frac{1}{30}$

v1 = 30 cm

This image would now act as a virtual object for 2nd lens.

u2 = +(30-8) = + 22 cm , v2 = ? f2 = -20cm

$\displaystyle \frac{1}{v_2} – \frac{1}{u_2} = \frac{1}{f_2}$

$\displaystyle \frac{1}{v_2} – \frac{1}{22} = \frac{1}{-20}$

v2 = -220 cm

Parallel incident beam would appear to diverge from a point 220 – 4 = 216 cm from the centre of the two lens system.

(ii) Suppose a parallel beam of light from the left is incident first of the concave lens.

u1 = -∞ , f1 = -20 cm , v1 = ?

$\displaystyle \frac{1}{v_1} – \frac{1}{u_1} = \frac{1}{f_1}$

$\displaystyle \frac{1}{v_1} – \frac{1}{-\infty} = \frac{1}{-20}$

v1 = -20 cm

The image acts as a real object for the 2nd lens

u2 =-(20+8) =-28 cm, f2 = 30 cm, v2 =?

$\displaystyle \frac{1}{v_2} – \frac{1}{u_2} = \frac{1}{f_2}$

$\displaystyle \frac{1}{v_2} – \frac{1}{-28} = \frac{1}{30}$

v2 = -420 cm

The parallel beam appears to diverge from a point 420 – 4 = 416 cm, on the left of the centre of the two lens system.
From the above discussion, we observe that the answer depends on which side of the lens system the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be useful here.

Here, h1 = 1.5 cm, u1 =-40 cm , m = ?, h2= ?

For the 1st lens ;

$\displaystyle \frac{1}{v_1} – \frac{1}{-40} = \frac{1}{30}$

v1 = 120 cm

Magnitude of magnification produced by first lens;

$\displaystyle m_1 = \frac{v_1}{u_1}$

$\displaystyle m_1 = \frac{120}{40}= 3$

The image formed by 1st lens acts as virtual object for the 2nd lens

u2=120-8=112 cm, f2 = -20 cm, v2= ?

$\displaystyle \frac{1}{v_2} – \frac{1}{u_2} = \frac{1}{f_2}$

$\displaystyle \frac{1}{v_2} – \frac{1}{112} = \frac{1}{-20}$

$\displaystyle v_2 = \frac{-112 \times 20}{92}$

Magnitude of magnification produced by second lens ;

$\displaystyle m_2 = \frac{v_2}{u_2}$

$\displaystyle m_2 = \frac{112 \times 20}{92 \times 112} = \frac{20}{92}$

Net magnification produced by the combination m = m1 × m2

$\displaystyle m = 3 \times \frac{20}{92}$

m = 0.652

size of image, h2 = m × h1 = 0.652 × 1.5 = 0.98 cm

Q: 22. At what angle should a ray of light be incident on the face of a prism of refracting angle 60°, so that it just suffers total internal reflection at the other face? The refractive index of the prism is 1.524.

Sol: i1 = ?, A = 60° , μ = 1.524

$\displaystyle \mu = \frac{1}{sinC}$

C = r2 (for total internal reflection at face AC)

sin C = sin r2 = 1/μ = 1/1.524 = 0.656

r2 = 41°

As r1 + r2 = A

r1 = A – r2 = 60° – 41° = 19°

As , μ = sin⁡i1/sin⁡r1

sin i1 = μ sin r1

sin i1 = 1.524 sin19°

sin i1 = 1.524 × 0.3256 = 0.4962

i1 = sin-1 (0.4962) = 29° 45′

Q:23. A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a)What is the angular 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

Sol: Here, area of each (object) square = 1 mm2 , u = – 9 cm, f = 10 cm.

As, $\displaystyle \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

$\displaystyle \frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

$\displaystyle \frac{1}{v} = \frac{1}{10} – \frac{1}{9}$

1/v = -1/90

v = -90 cm

Magnification, m = v/|u| = 90/9 = 10

Area of each square in virtual image = (10)^2×1=100 sq.mm

Magnifying power = d/u = 25/9 = 2.8

No, magnification in (a) which is (v/u) cannot be equal to magnifying power in (b) which is (d/u) unless v = d i.e. image is located at the least distance of distinct vision.

Q:24. (i) At what distance should the lens be held from the figure in Q.23 in order to view the square distinctly with maximum possible magnifying power ?

(ii) What is the magnification in this case ?

(iii) Is the magnification equal to magnifying power in this case ? Explain .

Solution : (i) v = -25 cm , f = 10 cm , u = ?

$\displaystyle \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

$\displaystyle \frac{1}{-25} – \frac{1}{u} = \frac{1}{10}$

u = -50/7 = -7.14 cm

(ii) Magnification , $\displaystyle m = \frac{v}{|u|}$

$\displaystyle m = \frac{25}{7.14}$

m = 3.5

(ii) Magnifying Power , $\displaystyle = \frac{d}{u}$

$\displaystyle = \frac{25}{7.14}$

Here , the magnifying power & magnification in this case are equal , because image is formed at least distance of distinct vision .

Q:25. What should be the distance between the object (in Q.24) and magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2 . Would you be able to see the squares distinctly with your eyes very close to the magnifier ?

Sol: Area magnification = 6.25 mm2

Linear magnification , $\displaystyle m = \sqrt{6.25}$

m = 2.5

As , $\displaystyle m = \frac{v}{u}$

v = m u = 2.5 u

$\displaystyle \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

$\displaystyle \frac{1}{2.5 u} – \frac{1}{u} = \frac{1}{10}$

u = – 6 cm

v = 2.5 u = 2.5 × (-6) = -15 cm

As the virtual image is at 15 cm; whereas distance of distinct vision is 25 cm, therefore, the image cannot be seen distinctly by the eye.

(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 eye very close to the lens. Does angular magnification change if the eye is moved back?

(c)Magnifying power of 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 eye piece of a compound microscope have short focal lengths ?

(e)When viewing through a compound microscope, our eyes should be positioned not on the eye piece, but a short distance away from it for best viewing, why? How much should be that short distance between the eye and eye piece?

Sol: (a) It is true that angular size of image is equal to angular size of the object.
By using magnifying glass, we keep the object far more closer to the eye than at 25 cm, it normal position without use of glass. The closer object has larger angular size than the same object at 25 cm. it is in this sense that angular magnification is achieved.

(b)Yes, the angular magnification changes, if the eye is moved back. This is because angle subtended at the eye would be slightly less than the angle subtended at the lens. The effect is negligible when image is at much larger distance.

(c)Theoretically, it is true. However, when we decrease focal length, aberrations both spherical and chromatic become more pronounced. Further, it is difficult to grind lenses of very small focal lengths.

(d)Angular magnification of eye piece is (1 + d/fe ). This increases as fe decreases.
Further, magnification of objective lens is v/u. As object lies close to focus of objective lens u ≈ fo. To increase this magnification (v/fo),fo should be smaller.

(e)The image of objective lens in eye piece is called ‘eye ring’. All the rays from the object refracted by the objective go through the eye ring. Therefore, ideal position for our eyes for viewing is this eye ring only. When eye is too close to the eye piece, field of view reduces and eyes do not collect much of the light. The precise location of the eye ring would depend upon the separation between the objective and eye piece, and also on focal length of the eye piece.

Q:27. An angular magnification (magnifying power) of 30 X is desired using and objective of focal length 1.25 cm and an eye piece of focal length 5 cm. How will you set up the compound microscope ?

Sol: In normal adjustment, image is formed at least distance of distinct vision, d = 25 cm.

Angular magnification of eye piece $\displaystyle m_e = 1 + \frac{d}{f_e}$

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

me = 6

As total magnification is 30,

m = mo × me

magnification of objective lens,

mo = 30/6 = 5

⇒ mo = vo/(-uo) = 5

or, vo = -5uo

As, $\displaystyle \frac{1}{v_0} – \frac{1}{u_o}=\frac{1}{f_0}$

$\displaystyle \frac{1}{-5u_0} – \frac{1}{u_o}=\frac{1}{1.25}$

6/(-5uo ) = 1/1.25

uo = (-6×1.25)/5 = -1.5 cm

i.e., object should be held at 1.5 cm in front of objective of lens.

As vo = -5uo

⇒ vo = -5 (-1.5) = 7.5 cm

From, $\displaystyle \frac{1}{v_e} – \frac{1}{u_e}=\frac{1}{f_e}$

$\displaystyle \frac{1}{-25} – \frac{1}{u_e}=\frac{1}{f_e}$

$\displaystyle \frac{1}{-25} – \frac{1}{5}=\frac{1}{u_e}$

1/ue = -6/25

ue = (-25)/6 = -4.17 cm

Separation between the objective lens and eye piece =|ue |+|vo

= 4.17 + 7.5 = 11.67 cm

Q:28. A small telescope has an objective lens of focal length 140 cm and eye piece of focal length 5.0 cm. what is the magnifying power of 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).

Sol: Here, fo = 140 cm , fe = 5.0 cm

Magnifying power = ?

In normal adjustment, Magnifying power $\displaystyle M = -\frac{f_0}{f_e}$

$\displaystyle M = -\frac{140}{5} =-28$

When final image is at the least distance of distinct vision,

Magnifying power, $\displaystyle M = -\frac{f_0}{f_e}(1+\frac{f_e}{d})$

$\displaystyle M = -\frac{140}{5}(1+\frac{5}{25})$

M = -33.6

Q:29. (a) For the telescope described in Q .28, what is the separation between the objective lens and eye piece ?

(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 ?

Separation between objective lens and eye lens= fo + fe

= 140 + 5 = 145 cm

α = 100/(3×1000) = 1/30 radian

If h is the height of image formed by the objective,

then α = h/fo = h/140

h/140 = 1/30

or, h= 140/30 cm = 4.7 cm

Magnification produced by eye piece = (1 + d/fe )= 1 + 25/5=6

h/140 = 1/30

or, h = 140/(30)= 4.7 cm

Magnification produced by eye piece =(1 + d/fe )= 1 + 25/5=6

Height of final image = 4.7 × 6 = 28.2 cm

Q:30. A Cassegrainian telescope uses two mirrors as shown in Figure (Textbook).  Such a telescope is built with the mirrors 20 mm apart. If radius of curvature of larger mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be ?

Sol: Here, radius of curvature of objective mirror

R1 = 220 mm

Radius of curvature of secondary mirror

R2 = 140 mm;

f2 = R2/2 = 140/2=70 mm

Distance between the two mirrors, d = 20 mm. When object is at infinity, parallel rays falling on objective mirror, on reflection, would collect at its focus at

f1 = R1/2 = 220/2 = 110 mm

instead, they fall on secondary mirror at 20 mm from objective mirror

For secondary mirror, u = f1 – d = 110-20 = 90 mm

From, $\displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f_2}$

$\displaystyle \frac{1}{v} = \frac{1}{f_2}-\frac{1}{u}$

$\displaystyle \frac{1}{v} = \frac{1}{70}-\frac{1}{90}$

1/v = 2/630

v = 630/2 = 315 mm

v =31.5 cm to the right of secondary mirror.

Q:31. Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Figure . A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away ?

Sol: 2θ = 2 × 3.5° = 7° = 7π/180 rad

From Fig. tan⁡2θ = SA/OS = d/1.5

d = 1.5 tan2θ ≈ 1.5(2θ)

= 1.5×7π/180 m = 0.18 m

Q:32. Figure (in Textbook) shows an equiconvex lens (of refractive index 1.5) in contact with a liquid layer on top of a lane 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?

Sol: Let focal length of convex lens of glass = f1 = 30 cm

focal length of Plano concave lens of liquid = f2

Combined focal length, F = 45.0 cm

As, $\displaystyle \frac{1}{F}= \frac{1}{f_1} + \frac{1}{f_2}$

$\displaystyle \frac{1}{f_2} = \frac{1}{F}- \frac{1}{f_1}$

$\displaystyle \frac{1}{f_2} = \frac{1}{45}- \frac{1}{30}$

$\displaystyle \frac{1}{f_2} = – \frac{1}{90}$

f2 = -90 cm

For glass lens, let R1 = R , R2 = – R

As , $\displaystyle \frac{1}{f_2} = (\mu_l-1) (\frac{1}{R_1}-\frac{1}{R_2})$

$\displaystyle \frac{1}{f_2} = (\mu_l-1) (\frac{1}{-30}-\frac{1}{\infty})$

$\displaystyle \frac{1}{-90} = (\mu_l-1) (\frac{1}{-30})$

μl-1= 30/90=1/3

μl = 1 + 1/3 = 4/3