**NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Solutions**,

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**NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments**

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**Ray Optics and Optical Instruments**

**Ray Optics and Optical Instruments****Chapter: 9**

Part – II |

**EXERCISE**

**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? 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?**

Ans: **Given:**

1/f = 1/do + 1/di

**Rearranging the equation to solve for di:**

1/di = 1/f – 1/do

= 1/18 – 1/27

= (3-2) / 54

= 1/54

di = 54 cm

**Size of the image:**

Magnification = – di/ do = – 54 / 27 = -2

h’ = m × h

= – 2 × 2.5 cm

= – 5 cm

**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.**

Ans: **Given:**

h = 4.5 cm

u = – 12 cm

f = 15 cm

**Here:**

1/f = 1/v + 1/u

1/15 = 1/v + 1 / (-12)

1/v = 3/20

20/3 = 6.67 cm.

**Magnification:**

m = -v/u = – 6.67 / (-12)

= 0.56.

**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 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?**

Ans: **Given:**

Real depth = 12.5 cm

Apparent depth = 9.4 cm

n = Real depth / Apparent Depth

n = 12.5 cm / 9.4 cm

= 1.33.

**Effects on replacing:**

d’ = real depth / n’

= 12.5 cm / 1.63

= 7.67 cm

**Distance to move the microscope:**

= 9.4 cm – 7.67 cm

= 1.73 cm.

**4. Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° 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° with the normal to a water-glass interface [Fig. 9.27(c)].**

Ans:

= From Eps. (i) and (ii), we get:

sin r = 1.32 × 0.7071 / 1.51

= 0.6181.

**5. A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. 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.)**

Ans:** Given:**

n_{Water} = 1.33

n_{Air} = 1

Sin (θ_{c}) = n_{air} / n_{water}

= 1 / 1.33 = 0.75

**Critical angle:**

θ_{c} = sin-1 (0.75)

= 48.75°

**Determine the radius:**

h = 80 cm

r = h × tan (θ_{c})

= 80 × tan (48.75°) = 91.2 cm

**Area:**

A = 𝜋r^{2} = 𝜋 × (91.2)^{2}

= 26, 144 cm^{2}.

**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°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.**

Ans:

**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 20cm?**

Ans: **Given:**

1/𝑓 = (n-1) (1/R_{1} – 1/R_{2})

(R_{1} = R and R_{2} = – R)

1/𝑓 = (n – 1) [1/R – (1/-R)]

1/𝑓 = 2(n -1) / R

R = 2 (n – 1)𝑓 / 1

**Substituting the given values:**

n = 1.55

𝑓 = 20 cm

R = 2 × (1.55 – 1) × 20 / 1

= 2 × 0.55 × 20 / 1

= 22 cm.

**8. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is:**

**(a) A convex lens of focal length 20cm, and.**

Ans: **Given:**

1/𝑓 = 1/v – 1/u

(u = -12 cm)

𝑓 = 20 cm

1/v = 1/𝑓 + 1/u

= 1/20 + 1/-12

1/v = -12 + 20 / 240

8/240 = 1/30

v = 30 cm.

**(b) A concave lens of focal length 16cm?**

Ans: **Given:**

𝑓 = -16 cm

**Here:**

1/v = 1/𝑓 + 1/u = (1/-16) + (1/ -12)

= 1/v = -12 -16 / 192

= -28/ 192

= -7 / 48

= – 6.86 cm.

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

Ans: **Given:**

h = 3.0 cm

u = – 14 cm

f = -21 cm

**Here:**

1/f = 1/v + 1/u

1/ (-21) = 1/v + 1(-14)

1/v = 1/ (-21) + 1/14

1/v = 1/42

v = 42 cm

m = -v/u = – 42 /(-14) = 3

h = m × h = 3 × 3.0 = 9cm.

The image is virtual, erect, magnified, and 9 cm in size. It is formed 42 cm behind the lens. If the object is moved further away, the image will move closer to the lens and become smaller.

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

Ans: **Given:**

𝑓_{1} = 30 cm

𝑓_{2} = – 20 cm

1/F = 1/𝑓_{1} + 1/𝑓_{2}

= 1/30 + 1/(-20)

= 1/30 – 1/20

= 2 – 3/ 60

= – 60 cm.

**11. A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. 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 (25cm). and**

Ans: **Here:**

𝑓_{0} = 2.0 cm

𝑓_{e} = 6.25 cm

L = 15 cm

(v_{e} = – D = – 25 cm)

1/v_{e} – 1/u_{e} = 1/f_{e}

i.e., 1/(-25) – 1/u_{e} = 1/ 6.25

Therefore – 1/u_{e} = 1/ 6.25 + 1/25 = ⅕

u_{e} = – 5 cm

v0 = L – lu_{e}l = 15 – 5 = 10 cm

**(b) At infinity? What is the magnifying power of the microscope in each case?**

Ans: v_{0} = d – 𝑓_{e} = 15 – 6.25 = 8.75 cm

1/v_{0} – 1/u_{0} = 1/f_{0}

1/ 8.75 – 1/ u_{0}

= ½ = – 2.59 cm.

**Magnifying power:**

M = ((8.75 / (-2.59)) × 4 = – 13.5.

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

Ans: **Given:**

𝑓_{0} = 8.0 mm = 0.8 m

𝑓_{e} = 2.5 cm

u_{0} = 9.0 mm = 0.9 cm

D = 25 cm

**Here:**

1/v_{0} – 1/u_{0} = 1/𝑓_{0}

1/v_{0} – 1/u_{0} = 1/0.8

1/v_{0} = 1/0.8 + 1/0.9

= 1.25 + 1.11

= 2.36 cm^{-1}

v_{0} = 0.42 cm.

**Separation between the two lenses (d):**

d = v_{0} + u_{e}

(v_{e} = – 25 cm)

1/v_{e} – 1/u_{e} = 1/𝑓_{e}

d = 7.2 cm + 2.27 cm

= 9.47 cm.

M = 8 × (1+10) = 8 × 11 = 88.

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

Ans: **Given:**

𝑓_{0} = 144 cm

𝑓_{e} = 6.0 cm

m = 𝑓_{0} / 𝑓_{e}

= 144/ 6.0

= 24 cm

d = 𝑓_{0} + 𝑓_{e}

= 144 cm + 6.0 cm

= 150 cm.

**14. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm is used, what is the angular magnification of the telescope?**

Ans: **Given:**

𝑓_{0} = 15 m

𝑓_{e} = 1.0 cm = 0.01 m

M = 𝑓_{0} / 𝑓_{e}

= 15 / 0.01

= 1500.

**(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 × 10**^{6}**m, and the radius of lunar orbit is 3.8 × 10**^{8}**m.**

Ans: **Given:**

D = 3.48 × 10^{6} m

r = 3.8 × 10^{8} m

α = D/r = 3.48 × 10^{6} / 3.8 × 10^{8}

= 13.7 × 10^{-2} m or 13.7 cm.

**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.**

Ans: Object distance between 𝑓 and 2𝑓.

Suppose 𝑓 < do < 2𝑓.

1/𝑓 = 1/d0 + 1di

= 1/di = 1/𝑓 – 1/d0

**Determine di:**

Since d0 > 𝑓:

1/𝑓 – 1/d0 > 0

di = 𝑓 × d0 / d0 – 𝑓

Because d0 is between 𝑓 and 2𝑓

d0 – 𝑓 is positive and less than 𝑓

**(b) A convex mirror always produces a virtual image independent of the location of the object.**

Ans: From mirror formula v = uf/u-f.

For a convex mirror, the focal length 𝑓 is negative and the object distance d_{o} is positive. Since d_{i} is always negative, the image is always virtual and formed behind the mirror. This virtual image is always diminished in size.

**(c) The virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.**

Ans: From relation, m = 𝑓 / u – 𝑓 and 𝑓 being positive for convex mirror, m will always be negative and less than one. Hence virtual image formed will also be diminished.

From relation, m = v-𝑓 / 𝑓 and m being negative, v will always be less than, 𝑓 Hence image will be formed between pole and focus.

**(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.]**

Ans: **When u > 0 < 𝑓 , we get:**

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

Ans: **Given:**

t = 15 cm

n = 1.5

d = t (1 – 1/n)

d = 15 × (1 – 1/1.5)

= 15 × (1 – 2/3)

= 15 × 1/3

= 5 cm.

**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.**

Ans: **Given:**

n^{1} = 1.68

n^{2} = 1.44

**Total internal reflection condition:**

Sin θc = n^{2}/ n^{1}

= 1.44 / 1.68

= 0.857

θc = sin^{-1} (0.857) = 59.0°

**Angle with the Axis of the pipe:**

Sin θ = cos *i*

Since θ > θc:

cos*i* > sin 59.0° = cos *i* > 0.857

i < cos^{-1} (0.857) = 31.0

0 < i < 31.0°.

**(b) What is the answer if there is no outer covering of the pipe?**

Ans: **Given: **

n^{2} = 1.00.

Sinθ_{c} = n^{2}/n^{1} = 1.00/1.68 = 0.595.

θ_{c} = sin^{-1} (0.595) = 36.5°

For TIR to occur θ> 36.5°

cos* i* > sin 36.5 = cos* i* > 0.595

*i* < cos-1 (0.595) = 52.0°.

**18. 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?**

Ans: **Given:**

1/𝑓 = 1/d_{0} + 1/d_{i}

= d_{0} + d_{i} = 3 cm

d_{i} = 3 – d_{0}

1/𝑓 = 1/d_{0} + 1/ 3 – d_{o}

(d_{o} = d_{i} = 3/2 m)

= 𝑓 = 3/4 m

= 0.75 m.

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

Ans: **Given:**

D = 90 cm

d = 20 cm

𝑓 = D^{2} – d^{2 }/ 4D

= 90^{2} – 20^{2} / 4 × 90

= 8100 – 400 / 360

= 7700 / 360

= 21.39 cm.

**20. (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm 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?**

Ans: **Given:**

𝑓_{1} = 30 cm

𝑓_{2} = – 20 cm

d = 8.0 cm

1/F = 1/𝑓_{1} + 1/𝑓_{2} – d/ 𝑓_{1}𝑓_{2}

1/F = 1/30 + 1/(-20) – 8.0 / (30) (-20)

= 1/30 – 1/20 + 8.0 / 600

= 1/30 – 1/20 + 1/75

= 2/60 – 3/60 + 4/300

= (-1) / 60 + 4/300

**Convert 4/300 to a common denominator:**

= 1/F = – 1 / 60

= 60 cm.

**(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm. Determine the magnification produced by the two-lens system, and the size of the image.**

Ans: **Given:**

u_{1} = 40 cm

𝑓_{1} = 30 cm

1/v_{1} – 1/u_{1} = 1/f_{1}

1/v_{1} + 1/40 = 1/30

= 1/v_{1} = 1/30 – 1/40 = 4-3/ 120

= 120 cm

= (-112) + 20 / 112 × 20

= – 92 / 112 × 20

v_{2} = 112 × 20 / 92

= – 24.9 cm

m = m_{1} × m_{2} = 3 × 20 / 92

= 0.652

Size of image:

h_{2} = nh_{1} = 0.652 × 1.5

= 0.98 cm.

**21. 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 material of the prism is 1.524.**

Ans:** Given:**

A = 60°

n = 1.524

= sin θ_{c} = 1/n

Sin θ_{c} = 1/1.524 = 0. 656

θ_{c} = sin^{-1} (0.656) = 41.14°

**Use the geometry of the prism:**

r_{1} + r_{1} = A

r_{1} = A – θc

r_{1} = 60° – 41.14°

= 18.86°

**Apply snell’s law at the first face:**

Sin *i* = n sin r_{1}

= 1.524 × sin 18.86°

= 1.524 × 0.324 = 0.494

*i* = sin ^{-1} (0.494) = 29.53°.

**22. 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 9 cm) held close to the eye.**

**(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?**

Ans: **Given:**

𝑓 = 9 cm

u = – 9 cm

The area of each square on the card sheet is 1 mm^{2}.

**Here:**

1/𝑓 = 1/v – 1/u

= 1/9 = 1/v – 1/(-9)

= 1/v = 1/9 + 1/9

= 2/9

v = 9/2 = 4.5 cm

**Magnification (M):**

M = -v/u = (- 4.5) / (-9)

= 1/2.

**(b) What is the angular magnification (magnifying power) of the lens?**

Ans: **Angular Magnification:**

Mp = 1 + D/ 𝑓

= 1 + 25 / 9

= 1 + 2.78

= 3.78.

**(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.**

Ans: No. Magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. Magnification and angular magnification are complementary concepts that describe different aspects of image formation through optical instruments. Understanding both is essential for appreciating the capabilities and limitations of various lenses and instruments.

**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? **

Ans: **Given:**

𝑓 = 9 cm

D = 25 cm

(v = -D = – 25 cm)

**Here:**

1/𝑓 = 1/v – 1/u

1/9 = 1/(-25) – 1/u

= 1/u = 1/ (-25) – 1/9

= (-9) + 25 / 225

= 16/225

= 14.06 cm.

**(b) What is the magnification in this case?**

Ans: **Given:**

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

Ans: **Given:**

Mp = 1 + D/𝑓

= 1 + 25/9

= 3.78.

**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 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.]

Ans: **Given:**

𝑓 = 9 cm

A: 6.25 mm^{2}

**Magnification:**

M = √6.25 / 1

= √6.25 = 2.5

Distance u =

u = (M – 1) × 𝑓 / M

= (2.5 – 1) × 9 cm/ 2.5

= 5.4 cm.

**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?**

Ans: Since the size of the image produced by the magnifying glass is much larger than the size of the object, and the angular size of the image is equal to the angular size of the object, the magnifying glass helps to view objects that are closer than the standard least distance of distinct vision (approximately 25 cm). The closer the object is to the magnifying glass, the larger its angular size appears.

**(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?**

Ans: Indeed, the angular magnification changes with the distance between the eye and the magnifying glass. As the distance increases, the angular magnification decreases.

**(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?**

Ans: We cannot make lenses having very small focal length very easily. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.

**(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?**

Ans: The objective lens must have a short focal length to produce a highly magnified real image of the object. The eyepiece also needs a short focal length to further magnify this real image produced by the objective.

**(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?**

Ans: When we position our eye too close to the eyepiece of a compound microscope, we may not be able to collect a sufficient amount of refracted light because the field of view becomes limited. This reduction in the field of view can lead to a blurred image and decreased clarity.

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

Ans: **Given:**

D = 25 cm

𝑓_{e} = 5 cm

m_{e} = 1 + D/𝑓 = 1 + 25/5 = 5 cm

m_{o} = m/m_{e} = 30/6 = 6

m_{o} = v_{o}/u_{o} = -5

v_{o} = -5 uo

𝑓_{o} = 1.25 cm

**Now:**

1/vo – 1/u_{o} = 1/𝑓o

1/-5u_{o} – 1/u_{o} = 1/ 1.25

-6 / 5u_{o} = 1/ 1.25

u_{o} = 6 × 1.25 / 5 = – 1.5 cm.

v_{o} = -5 u_{o} = – 5 (-1.5 ) = 7.5 cm

1/v_{e} – 1/u_{e} = 1/f_{e}

= 1/v_{e} – 1/f_{e} = 1/(-25) – ⅕

= (-1 – 5) / 25

= – 6/25

u_{e} = 25 / 6

= – 4.17 cm

= 11.67 cm.

**27. A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. 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 (25cm)?**

Ans: M = 𝑓_{o} / 𝑓_{e}

= 140 cm / 5.0 cm

= 28 cm.

Magnifying power with Final image at 25 cm

(Me): = 1 + 25/5.0 = 6

Mtotal = 𝑓_{o}/ 𝑓_{e} × M_{e} = 28 × 6 = 168 cm.

**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 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 25cm?**

Ans: **Separation:**

140 cm + 5 cm = 145 cm

Magnification_{objective}:

M = -v / u = h’/h

h’ = -v × h / u

= – 140.7 cm × 100 m / 3000 m

= – 4.69 cm

**Height of the final image is formed at 25 cm:**

M = 𝑓_{o} / 𝑓_{e}

= 140 cm / 5 cm

= 28

M_{total} × h = 28 × 100m = 2800 cm.

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

Ans: **Given:**

𝑓_{1} = 110 mm

𝑓_{2} = 70 mm

d = 20 mm

**Here:**

v_{2} = (- 𝑓_{1}) – d

= -(110 mm – 20 mm)

= – 90 mm.

1/v_{2} = 1/𝑓 – 1/u_{2} + 1/v_{2}

= 1/70 + 1/90

1/v_{2} = (90 + 70) / (70 × 90)

u_{2} = 39.4 mm

u_{1} = v_{2} – d

20 mm – 39.4 mm = -19.4 mm

1/v_{1} = 1/𝑓_{1} – 1/u_{1} + 1/v_{1}

= 1/110 – 1/(-19.4) 1/v_{1}

= 19.4 + 110 / 110 × 19.4 1/v_{1}

= 129.4 / 2134

v_{1} = 16.5 mm.

**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 3.5o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?**

Ans: When a mirror rotates by an angle θ, the reflected ray rotates by 2θ,

So, the angle of deflection of the reflected ray is 2 × 3.5°

= 7°.

**Given:**

tan (7°) = d/1.5 m

d = 1.5 m × tan (7°)

= 18.4 cm.

**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 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?**

Ans: **Given:**

1/𝑓 = (μl – 1) × (1/R_{1} – 1/R_{2})

R_{1} = (-R_{2}) = R

So 1/𝑓 = (μl – 1) × (2/R)

Given 𝑓 = 45.0 cm

R = 2 × (μl – 1) × 𝑓

= 2 × (1.50 – 1) × 45.0 cm

= 45.0 cm

Focal length without liquid (𝑓)

(μl – 1)

1/𝑓 = (1 – 1) × (2/R) = 0

**Effective focal length:**

1/feff = 1/f’ – 1/f

f is infinite, 1/f = 0.so, 1/feff = 1/f

feff = f’ = 45.0 cm.

**Refractive index of the liquid (μ**_{2}**):**

1/feff = (μl – μ_{2}) × (1/R_{1} – 1/R_{2})

1/45.0 cm = (1.50 – μ_{2}) × (2/45.0 cm)

μ_{2} = 1.50 – 1 = 0.50