Ncert Solution Class 12th Physics Chapter – 9 Ray Optics Optical Instruments Question & Answer

NCERT Solutions Class 12th Physics Chapter – 9 Ray Optics Optical Instruments

NCERT Solutions Class 12th Physics Chapter – 9 Ray Optics Optical Instruments

?Chapter – 9?

✍Ray Optics Optical Instruments✍

?Question & Answer?

Question 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?

Answer: Given: The size of the candle (object) is 2.5 cm, the distance of the candle (object) from the mirror is 27 cm and the radius of curvature is 36 cm.

The formula of the mirror is given as,

1 u + 1 v = 1 f 1 u + 1 v = 1 ( R 2 )       [ ∵f= R 2 ] 1 u + 1 v = 2 R

Where, the distance of the object is u, the distance of the image is v, the radius of curvature is R and the focal length is f.

By substituting the values in the above formula, we get

1 −27 + 1 v = 2 −36 1 v = 1 −18 − 1 −27 1 v =− 1 54 v=−54 cm

The negative sign indicates that the image is formed in front of the mirror.

Thus, the image is formed at 54 cm at the front of the mirror.

The magnification is given as,

m= −v u h i h o = −v u

Where, the object height is h o and the image height is h i .

By substituting the values in the above expression, we get

h i 2.5 =−( −54 −27 ) h i =−5 cm

Thus, the image formed is inverted and real. As, the candle is moved closer to mirror the screen would have to be moved farther.

When the candle closer than 18 cm from the mirror, the image would be virtual and therefore cannot be collected on the screen.

Question 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.

Answer:

Height of the needle, h₁ = 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:

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 is given by the magnification formula:

Hence, magnification of the image, m =h₂ / h₁=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.

Question 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?

Answer:

Question 4:

The figures above show the 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.

As per the given figure, for the 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:

As per the given figure, for the 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:

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

The following figure shows the situation involving the glass − water interface.

Hence, the angle of refraction at the water − glass interface is 38.68°.

Question 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.)

Answer: Actual depth of the bulb in water, d₁ = 80 cm = 0.8 m

Refractive index of water, μ = 1.33

The given situation is shown in the following figure:

Where,

i = Angle of incidence

r = Angle of refraction = 90°

Since the bulb is a point source, the emergent light can be considered as a circle of radius, R = AC/2 = AO = OB

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

μ = Sin r / Sin i

1.33 = Sin 90° / Sin i

∴ i = Sin^-1(1/1.33) = 48.75°

Using the given figure, we have the relation:

tan i = OC/OB = R/d₁

∴ R = tan 48.75° × 0.8 = 0.91 m

∴ Area of the surface of water = πR² = π (0.91)² = 2.61 m²

Question 6.  A prism is made of glass of unknown refractive index. A parallel beam of light is incident on the 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.

Answer:

Angle of minimum deviation, δ_m = 40°

Angle of the prism, A = 60°

Refractive index of water, µ = 1.33

Refractive index of the material of the prism = µ’

The angle of deviation is related to refractive index (µ’) as:

Hence, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water, let δ’_m‘ be the new angle of minimum deviation for the same prism.

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

μwgμ’μAδ’mA

Hence, the new minimum angle of deviation is 10.32°.

Question 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?

Answer:

Lens maker s formula,  1/f​ = (μ−1) (1/R₁​​−1/R_2​​)
Here, $f=20cm,\mu =1.55$,
R₁​=R and R₂​=−R
∴1/20​=(1.55−1)(1/R−1/(−R)1​)or
1/20​=0.55×2/R​
$\Rightarrow R=1.1\times 20$
$=22cm$

 Question 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 (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16 cm?

Answer:

$(a)$. According to the given case, object is virtual, $u=+12cm$

Question 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? 

Answer:

Size of the object, h₁ = 3 cm

Object distance, u = −14 cm

Focal length of the concave lens, f = −21 cm

Image distance = v

According to the lens formula, we have the relation:

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

Hence, the height of the image is 1.8 cm.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

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

Answer:

Focal length of the convex lens, f₁ = 30 cm 

Focal length of the concave lens, f₂ = −20 cm 

Focal length of the system of lenses = f 

The equivalent focal length of a system of two lenses in contact is given as:

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

Question 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 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer: Focal length of the objective lens (f₁) = 2.0 cm

Focal length of the eyepiece (f₂) = 6.25 cm

Distance between the objective lens and the eyepiece (d)= 15 cm

(a) Least distance of distinct vision, (d¹) = 25 cm

Therefore, Image distance for the eyepiece (v₂) = – 25 cm

Object distance for the eyepiece =  u₂

According to the lens formula,

1/ v₂ – 1/u₂ = 1/ f₂

1/u₂ = 1/ v₂ -1/ f₂  = 1/-25 – 1/6.25 = -1-4/ 25 = -5/25

 u₂ = -5 cm

Image distance for the objective lens, (v₁) =- d + u₂ = 15-5 = 10 cm

Object distance for the objective lens = u₂

According to the lens formula,

1/ v₁ – 1/u₁ = 1/f₁

1/u₁ = 1/ v₁ – 1/f₁

        = 1/10 – 1/2 = 1-5/10 = -4/10

u₁ = -2.5 cm

Magnitude of the object distance (u₁) = 2.5 cm

The magnifying power of a compound microscope

m = v₁/ u₁ (1 + d¹/f₂)

    = 10/2.5 (1 + 25/6.25)

    = 4 (1 +4) = 20

Hence, the magnifying power of the microscope is 20.

(b)The final image is formed at infinity.

Therefore, mage distance for the eyepiece (v₂) = ∞

Object distance for the eyepiece = u₂

According to the lens formula,

1/v₁ – 1/u₁ = 1/f₁

 1/u₁ =  1/v₁ -1/f₁

          =  1/8.75 – 1/20 = 2-8.75/ 17.5

u₁ = -17.5/6.75 = -2.59 cm

Magnitude of the object distance (u₁) = 2.59 cm

The magnifying power of a compound microscope

m = v₁/ u₁ (d¹/f₂)

     = 8.75/2.59 x 25/6.25 = 13.51

Hence, the magnifying power of the microscope is 13.51.

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

Answer:

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 for the eyepiece, v_e = −d = −25 cm

Object distance for the eyepiece = u_e

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

We can also obtain the value of the image distance for the objective lens (v_o) using the lens formula.

The distance between the objective lens and the eyepiece

= | u_e | + V_o

= 2.27 + 7.2

= 9.47 cm 

The magnifying power of the microscope is calculated as:

Hence, the magnifying power of the microscope is 88.

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

Answer: Given that the focal length of the objective lens is 144 cm and focal length of the eyepiece is 6 cm.

Let m be the magnifying power of the telescope, then

m= f o f e

Here, f o be the focal length of the objective lens and f e be the focal length of the eyepiece.

Substitute the values in the above expression.

m= 144 6 =24

Let d be the separation between the objective lens and eyepiece, then

d= f o + f e

Substitute the values in the above expression.

d=144+6 =150 cm

Hence, the value of magnifying power is 24 and the separation between the objective lens and eyepiece is 150 cm.

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

(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 \times 10^{6} m3.48×106m,and the radius of lunar orbit is 3.8 \times 10^{8}m3.8×108m.

Answer: Focal length of the objective lens, f_o = 15 m = 15 × 10² cm

Focal length of the eyepiece, f_e = 1.0 cm

(a) The angular magnification of a telescope is given as:

Hence, the angular magnification of the given refracting telescope is 1500.

(b) Diameter of the moon, d = 3.48 × 10⁶ m

Radius of the lunar orbit, r₀ = 3.8 × 10⁸ m

Let d’ 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, the diameter of the moon’s image formed by the objective lens is 13.74 cm

 Question 15:
Use the mirror equation to deduce 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) the virtual image produced by a convex mirror is always diminished in size
and is located between the focus and the pole.
(iv) 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.]

Answer:

Mirror equation is given by,

(a) For a concave mirror, f is negative, i.e., f < 0

For a real object i.e., which is on the left side of the mirror, 

For u between f and 2f implies that 1/u lies between 1/f and 1/2f

i.e.,   Implies, v is negative and greater than 2f. Therefore, image lies beyond 2f and it is real.

(b) Focal length is positive for convex mirror, i.e., f > 0.

For a real object on the left side of the mirror, u is negative.

That is, 

Since u is negative and f is positive so, 1/v should also be positive, so v must be positive.

Hence, image is virtual and lies behind the mirror.

(c)  Using the mirror equation, we have

That is, the image is enlarged.

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

Answer: Actual depth of the pin, d = 15 cm

Apparent depth of the pin = d’

Refractive index of glass, μ = 1.5

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

μdd’μ=d/d’

∴ d’ = d/μ

= 15/1.5

= 10 cm

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.

 Question 17.   (i) Figure below 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.
(ii) What is the answer if there is no outer covering of the pipe?

Answer:

(a) Refractive index of the glass fibre, μ₁ = 1.68

Refractive index of the outer covering of the pipe, μ₂ = 1.44

Angle of incidence = i

Angle of refraction = r

Angle of incidence at the interface = i’

The refractive index (μ) of the inner core − outer core interface is given as:

= 0.8571

∴ i’ = 59°

For the critical angle, total internal reflection (TIR) takes place only when i > i’, i.e., i > 59°

Maximum angle of reflection, r_max = 90° − i’ = 90° − 59° = 31°

Let, i_max be the maximum angle of incidence.

The refractive index at the air-glass interface, μ₁ = 1.68

We have the relation for the maximum angles of incidence and reflection as:

sin i_max = μ₁ sin r_max

= 1.68 sin 31°

= 1.68 × 0.5150

= 0.8652

∴ i_max = sin⁻¹ 0.8652 ≈ 60°

Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection

(b) If the outer covering of the pipe is not present, then:

Refractive index of the outer pipe, μ₁ = Refractive index of air = 1

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

= 36.5°

∴ i’ = 90° − 36.5° = 53.5°

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

Question 18.  Answer the following questions:
(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 underwater, looks obliquely at a fisherman standing on the bank of a lake.
Would the fisherman look 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?

Answer:

(i) Yes.

Real images can be obtained from plane and convex mirrors too. When the light rays converge at a point behind the plane or convex mirror, the object is said to be virtual. The real image of this object is obtained on the screen which is placed in the front of the mirror. This is where the real image is obtained.
(ii) No

To obtain a virtual image, the light rays must diverge. In the eyes, the convex lens help in converging these divergent rays at the retina. This is an example where the virtual image acts as an object for the lens to produce a real image.

(iii) The diver is in the water while the fisherman is on the land. Air is less dense when compared to the water as a medium. It is mentioned that the diver is viewing the fisherman. This explains that the light rays are travelling from the denser medium to the rarer medium. This means that the refracted rays move away from the normal making fisherman appear taller.

(iv) Yes; Decrease

The reason behind the change in depth of the tank when viewed obliquely is because the light rays bend when they travel from one medium to another. This also means that the apparent depth is less than the near-normal viewing.

(v) Yes

2.42 is the refractive index of the diamond while 1.5 is the refractive index of the ordinary glass. Also, the critical angle of the diamond is less than the glass. The sparkling effect of a diamond is possible because of the large cuts in the angle of incidence. Larger cuts ensures that the light entering the diamond is totally reflected from all the faces.

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

Answer:

Given: The distance between the object and the image is 3 m.

The maximum focal length of the convex lens to obtain real image is given as,

f max = d 4

Where, the maximum focal length of the convex lens is f max and the distance between the object and the image is d.

By substituting the given values in the above expression, we get

f max = 3 4 =0.75 m

Thus, the maximum possible focal length of the convex lens is 0.75 m.

Question 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.

Answer:

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

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

Focal length of the lens = f

Focal length is related to d and D as:

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

Question 21. 
(i) Determine the ‘effective focal length’ of the combination of the two lenses in
Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does
the answer depend on which side of the combination a beam of parallel light is incident?
Is the notion of effective focal length of this system useful at all?

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

Answer:

Focal length of the convex lens, f₁ = 30 cm

Focal length of the concave lens, f₂ = −20 cm

Distance between the two lenses, d = 8.0 cm

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

According to the lens formula, we have:

1/v_{1  ^{_  1/μ}1 ^{= 1/f}1}

Where,

_{^μ1 = Object distance = ∞}

v₁ = Image distance

1/v₁ = 1/30 – 1/_{∞ = 1/30}

The image will act as a virtual object for the concave lens.

Applying lens formula to the concave lens, we have:

1/v₂ – 1/_^μ2 = 1/f₂

Where,

_{^μ1 = Object distance}

= −(20 + d) = −(20 + 8) = −28 cm

v_{1 = Image distance}

1/v1 = 1/30 + 1/-28 = 14 – 15/420 = -1/420

∴ v₂ = – 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 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.

(b) Height of the image, h₁ = 1.5 cm

Object distance from the side of the convex lens, _{^μ1 ^{=  – 40 cm | μ}1^{| = 40 cm}} 

According to the lens formula:

1/v₁ – 1/_{^μ1 ^{= 1/f}1}

Where,

v₁ = Image distance

1/v1 = 1/30  + 1/-40 = 4-3/ 120 = 1/ 20

∴ v₁ = 120 cm

Magnification, m = v₁ / | _{^μ1 ^|}

= 120 /40 = 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 the lens formula:

1/v₂ – 1/_{^μ2 ^{= 1/f}2}

Where,

_{^μ2^{= Object distance}}

= +(120 − 8) = 112 cm.

_{^v2 ^{=  Image distance}}

1/v₂  = 1/-20 = 1/112 -112 + 20/ 2240 = -92/2240

∴ v2 = -2240/92 cm

Magnification,  m’ = |v₂/u₂|

= 2240/92 x 1/112 = 20/92

Hence, the magnification due to the concave lens is 20/92

The magnification produced by the combination of the two lenses is calculated as:

m x m’

= 3 x 20/92 = 60/92 = 0.652

The magnification of the combination is given as:

h₂/h₁ = 0.652

h₂ = 0.652 x h₁ 

Where,

h₁ = Object size = 1.5 cm

h₂ = Size of the image

∴h₂ = 0.652 x 1.5 = 0.98 cm  

Hence, the height of the image is 0.98 cm.

Question 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 material of the prism is 1.524.

Answer:The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure.

Angle of prism, ∠A = 60°

Refractive index of the prism, µ = 1.524

i₁ = Incident angle

r₁ = Refracted angle

r₂ = Angle of incidence at the face AC

e = Emergent angle = 90°

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

= 0.6562

∴ r₂ = sin⁻¹ 0.6562 ≈ 41°

It is clear from the figure that angle A = r₁ + r₂

∴ r₁ = A − r₂ = 60 − 41 = 19°

According to Snell’s law, we have the relation:

= 1.524 × sin 19°

= 0.496

∴ i₁ = 29.75°

Hence, the angle of incidence is 29.75°.

Question 24.  For a normal eye, the far point is at infinity and the near point of distinct vision is about
25 cm in front of the eye. The cornea of the eye provides a converging power of about
40 dioptres, and the least converging power of the eye-lens behind the cornea is about
20 dioptres. From this rough data estimate the range of accommodation (i.e., the
range of converging power of the eye-lens) of a normal eye.

Answer: Given: The least distance of distinct vision is 25 cm, far point of a normal eye is infinite, the converging power of the cornea is 40 D and the least converging power of the eye-lens is 20 D.

Power of the eye-lens is given as,

P= P c + P e

Where, the power of the eye-lens is P, the converging power of the cornea is P c and the least converging power of the eye-lens is P e .

By substituting the given values in the above expression, we get

P=40+20 =60 D

Power of the eye-lens is given as,

To focus an object at the near point, object distance ( u ) is −d=−25 cm.

f e =D =v

Where, the focal length of the eye-lens is f e ( 100 60  cm ), the distance between the corona and the retina is D and the image distance is v.

By substituting the given values in the above expression, we get

v= 100 60 = 5 3  cm

The lens formula is given as,

1 f ′ = 1 ν − 1 u

Where f ′ is the focal length of the lens.

By substituting the given values in the above expression, we get

1 f ′ = 3 5 + 1 25 = 15+1 25 = 16 25   cm −1

Power is denoted as P’.

P ′ = 1 f ′ ×100

By substituting the given values in the above expression, we get

P ′ = 16 25 ×100 =64 D

The power of the lens is given as,

P L = P ′ −P

By substituting the given values in the above expression, we get

Where, the power of the lens is P L .

P L =64-40 =24 D

Thus, the range of accommodation of the eye-lens is from 20 D to 24 D.

Question 25.  Does the human eye partially lose its ability of accommodation when it undergoes short-sightedness (myopia) or long-sightedness (hypermetropia)? If not, what might cause these defects of vision?

Answer: When a person is suffering from myopia or hypermetropia, the eye-lenses are used. Myopia is a condition when the eyeballs start to elongate from the front to the back. Hypermetropia is a condition when the eyeballs start to shorten. While presbyopia is a condition where the eyeballs loses its ability of accommodation.

Question 26. A myopic person has been using spectacles of power –1.0 dioptre for distant vision. During old age he also needs to use separatereading glass of power + 2.0 dioptres. Explain what may have happened.

Answer:  Given: A myopic person has been using spectacles of power −1.0 D and during old age he uses separate reading glasses of power +2 D.

The focal length of spectacles is given as,

f= 1 P

Where, P is the power of spectacles.

By substituting the given values in the above expression, we get

f= 1 −1× 10 −2 =−100 cm

So, the far point of the person is 100 cm. He may have a normal near point of 25 cm. When he uses his spectacles, any object placed at infinity produces virtual image at 100 cm. Hence he uses the ability of accommodation of the eye-lens to see the objects placed between 100 cmand 25 cm

During old age, he requires reading glasses of power +2 D. This is because the ability of accommodation of eye is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.

Question 27. A person looking at a cloth with a pattern consisting of vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Answer: Since, the person is able to see vertical lines more distinctly than horizontal lines, the refracting system (cornea and eye-lens) of the eye is not working in the same manner at different planes. This defect is known as astigmatism. The person’s eye has sufficient curvature in the vertical plane hence; sharp images of the vertical lines are formed on the retina. However, the curvature in the horizontal plane is not enough for images to form sharply at the retina so the horizontal lines appear blurred. To correct this defect, cylindrical lenses can be used.

Question 28. A child with normal near point (25 cm) reads a book with small size print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What would be the shortest and the longest distance at which the lens should be placed from the page so that the book can be read easily when viewing through the magnifying glass?

(b) What is the max and the mini angular magnification (magnifying power) possible using the above given simple microscope?

Answer: 

(i)

Given: the focal length of the magnifying glass is 5 cm and the least distance of vision is 25 cm .

The lens formula is given as

1 f = 1 v − 1 u

Where, f is the focal length of the magnifying glass, v is the least distance of vision and u is the object distance.

When the man keeps the magnifying glass close to the page, the images are formed at his near point. So,

v=−25 cm

By substituting the given values in the above expression,

1 5 = 1 −25 − 1 u 1 u = 1 −25 − 1 5 = −5−1 25 u=−4.167 cm

Thus, the closest distance at which the person can read the book is 4.167 cm.

When the man keeps the magnifying glass far from the page, the images are formed at the far point of the eye. So,

v=∞

By substituting the given values in the above expression,

1 u ′ = 1 ∞ − 1 5 =− 1 5 u ′ =−5 cm

Thus, the farthest distance at which the person can read the book is 5 cm.

(ii)

The angular magnification is given as,

α= d | u |

Where, d is the near point of the eye.

Maximum angular magnification occurs when the magnifying glass is placed close to the book.

So,

| u |=4.167 cm

By substituting the given values in the above expression,

α max = 25 4.167 =6

Minimum angular magnification occurs when the magnifying glass is placed at the farthest distance from the book.

So,

| u |=5 cm

By substituting the given values in the above expression,

α min = 25 5 =5

Thus, the values of maximum and minimum angular magnification are 6 and 5 respectively.

Question 29.  A large card divided into squares each of size 1 mm² is being viewed from a distance of 9 cm through a magnifying glass ( converging lens has a focal length of 9 cm) held close to the eye. Determine:

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

(b) the angular magnification (magnifying power) of the lens?

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

Answer:  Given: The size of each square is 1  mm 2 , the object distance is 9 cm and the focal length of magnifying glass is 10 cm .

(a)

The lens formula is given as,

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

Where, object distance is u, focal length of converging lens is fand the image distance is v.

By substituting the given values in the above expression, we get

$1/10=1/v-1/-9$

$1/v=1/10-1/9$= 9−10 /90

$v=-90cm$

Magnification is given as,

m=− v /u

By substituting the given values in the above expression,

$m=-(-90/9)=10$

Each side of the square is magnified 10 times. So, the area of each square in the image is given as,

$A‘=A\times (10{)}^2$

Where, the area of each square is A .

By substituting the given values in the above expression, we get

A=1  mm 2 ×100 =100  mm 2 =1  cm 2

Thus, the area of each square in the image is 1  cm 2 .

(b)

Magnifying power of the lens is given as,

m= d | u |

Where, d is the least distance for distinct vision.

By substituting the given values in the above expression, we get

m= 25 | −9 | = 25 9 =2.8

Thus, the magnifying power of the lens is 2.8.

(c) The magnification in (a) is not the same as the magnifying power in (b).

The magnitude of magnification produce by the magnifying glass is 10 while its magnifying power is 2 .The two quantities will be equal when the image is formed at the near point of the eye at 25 cm.

Question 30:

(a)Determine the distance in which the lens should be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power.

(b)Determine the magnification in the following situation.

(c) Find if the magnifying power is equal to magnification.

Explain.

Answer: Given: The size of each square is 1  mm 2 , the object distance is 9 cmand the focal length of magnifying glass is 10 cm.

(a)

The lens formula is given as,

1 f = 1 v − 1 u

Where, object distance is u, focal length of converging lens is fand the image distance is v

The maximum possible magnification is obtained when the image is formed at the near point of eye.

So,

v=−25 cm

By substituting the given values in the above expression, we get

1 u = 1 v − 1 f 1 u = 1 −25 − 1 10 = −2−5 50 u=−7.14 cm

Thus, to view the squares distinctly, the lens should be kept 7.14 cmaway from them.

(b)

Magnification is given as,

m=| v u |

By substituting the given values in the above expression, we get

m=| −25 50 7 | = 25×7 50 =3.5

Thus, in this case the magnification is 3.5 times.

(c)

Magnifying power of the lens is given as,

m= d | u |

Where, d is the least distance for distinct vision.

By substituting the given values in the above expression,

m= 25 | 50 7 | = 25×7 50 =3.5

Thus, the magnifying power of the lens is 3.5.

Since, in this case the image is formed at the near point of the eye at 25 cm, the magnifying power is equal to the magnitude of magnification.

Question 31. The virtual image of each square in the figure is to have an area of 6.25 mm². Find out, what should be the distance between the object in Exercise 9.30 and the magnifying glass?If the eyes are too close to the magnifier, would you be able to see the squares distinctly?

Answer: Area of the virtual image of each square, A = 6.25 mm²

Area of each square, A₀ = 1 mm²

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

= 2.5

But m = image distance (V) / Object distance (u)

∴ v = mu = 2.5 u …………(1)

Focal length of the magnifying glass, f = 10 cm

According to the lens formula, we have the relation:

And v = 2.5 u

= 2.5 × 6 = −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 distinctly.

Question 32 . Answer the following questions:

(a) An object subtends an angle at the eye which is equal to the angle subtended at the eye by the virtual image that is produced by a magnifying glass. Does the magnifying glass provide angular magnification? Explain.

(b)A person’s eyes are very close to the lens when he is viewing through a magnifying glass. Does angular magnification change if the eye is moved back?

(c) The focal length of the lens is inversely proportional to the magnifying power of a simple microscope. Why don’t we achieve greater and greater magnifying power by using a convex lens of smaller and smaller focal length?

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

(e) Our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing, when viewing from a compound microscope. Explain why? How much should be that short distance between the eye and eyepiece?

Answer: (a) The angular size of the image and of the object are equal. When the objects are placed at the least distance of distinct vision, they can be viewed with the help of magnifying glass. As the object gets closer, the angular size increases. A magnifying glass is used because it provides an angular magnification. Without magnification, it is difficult to place an object closer to the eye.

(b) Yes, the angular magnification changes. As the distance between the eye and the magnifying glass increases, the angular magnification decreases. This is because the angle subtended by the eye is less than that of the angle subtended by the lens. Angular magnification is independent of the image distance.

(c) Manufacturing lenses with small focal length is difficult. When the focal length is small, there are chances of production of spherical and chromatic aberrations. Therefore, there is no such reduction in the focal length of a convex lens.

(d) Angular magnification of eye-piece is [(25/f_e​)+1] (f_e​ in cm) which increases if f_e​ is smaller. Further, magnification of the objective

is given by V_o​​/∣u_o​∣ = 1/ (∣u_o​∣/f_o​)−1

which is large when ∣u_o​∣ is slightly greater than $f$. The microscope is used for viewing very close object. So ∣u_o​∣ is small, and so is $f$.​

(e) We are unable to collect much-refracted light when we place our eyes too close to the eyepiece of a compound microscope. As a result, there is substantial decrements in the field of view. Hence, the clarity of the image gets blurred. The eye-ring attached to the eyepiece gives the best position for viewing through a compound microscope. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

Question 33.  An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

Answer:

Focal length of the objective lens,fo= 1.25 cm 

Focal length of the eyepiece, fe = 5 cm

Least distance of distinct vision, d = 25 cm 

Angular magnification of the compound microscope = 30 X 

Total magnifying power of the compound microscope, m = 30

The angular magnification of the eyepiece is given by the relation:

The angular magnification of the objective lens (mo) is related to me as:

Applying the lens formula for the objective lens:

1/f_o = 1/v_o – 1/u_o

1/1.25 = 1/-5u_o – 1/u_o = -6/5u_o

∴uo = -6/5 × 1.25 = – 1.5 cm

And v_o = -5u_o

= 5 × (-1.5) = 7.5 cm

The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification. 

Applying the lens formula for the eyepiece:

1/v_e – 1/u_e = 1/f_e

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm

Question 34:  A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?

(b) the final image is formed at the least distance of distinct vision (25 cm)?

Answer:

(a) Focal length of the objective lens, f_o​=140 cm

Question 35: (a)For a telescope, what is the separation between the objective lens and the eyepiece?

(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

(c)What is the height of the final image of the tower if it is formed at 25 cm?

Answer:

Focal length of the objective lens, f_o = 140 cm

Focal length of the eyepiece, f_e = 5 cm

(a) In normal adjustment, the separation between the objective lens and the eyepiece = f_o + f_e = 140 + 5 = 145 cm

(b) Height of the tower, h₁ = 100 m

Distance of the tower (object) from the telescope, u = 3 km = 3000 m

The angle subtended by the tower at the telescope is given as:

θ= h₁/u

= 100/3000 =1/30 rad

The angle subtended by the image produced by the objective lens is given as:

θ = h₂/f_o = h₂/140 rad 

Where,

h₂ = Height of the image of the tower formed by the objective lens

1/3 = h₂/140

∴h₂ = 140/30 = 4.7 cm 

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

(c) Image is formed at a distance, d = 25 cm

The magnification of the eyepiece is given by the relation:

m = 1 + d/f_o 

= 1+ 25/5 = 1 + 5 = 6

Height of the final image = mh₂ = 6 × 4.7 = 28.2 cm

Hence, the height of the final image of the tower is 28.2 cm.

Question 36: A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Answer: The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.

Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of the objective mirror, R₁ = 220 mm

Hence, focal length of the objective mirror, f₁ = R₁/2 = 110 mm

Radius of curvature of the secondary mirror, R₁ = 140 mm

Hence, focal length of the secondary mirror, f₂ = R₂/2 = 140/2 = 70 min

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.

Hence, the virtual object distance for the secondary mirror, u = f₁ − d

= 110 − 20

= 90 mm

Applying the mirror formula for the secondary mirror, we can calculate image distance (v) as:

1/V + 1/V = 1/f₂

1/V = 1/f₂ – 1/u

= 1/70-1/90

= 9-7/630

= 2/630

∴ v = 6302 = 315 mm

Question 37: Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. 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?

Answer: Angle of deflection, θ = 3.5°

Distance of the screen from the mirror, D = 1.5 m

The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ = 7.0°

The displacement (d) of the reflected spot of light on the screen is given as:

tan 2θ = d/1.5

∴ d = 1.5 × tan 7° = 0.184 m = 18.4 cm

Hence, the displacement of the reflected spot of light is 18.4 cm.

Question 38: Figure 9.37 shows a biconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Answer: Focal length of the convex lens, f₁ = 30 cm

The liquid acts as a mirror. Focal length of the liquid = f₂

Focal length of the system (convex lens + liquid), f = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as:

1/f = 1/f₁ + 1/f₂

1/f₂ = 1/f – 1/f₁

= 145-130

= -190

∴ f₂ = −90 cm

Let the refractive index of the lens be μ₁ and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is −R.

R can be obtained using the relation:

1/f = (μ1-1) (1/R+1/- R)

1/30 = (1.5-1)(2/R)

∴ R = 30/0.5×2 = 30 cm

Let μ₂ be the refractive index of the liquid.

Radius of curvature of the liquid on the side of the plane mirror = ∞

Radius of curvature of the liquid on the side of the lens, R = −30 cm

The value of μ₂ can be calculated using the relation:

1/f₂ = (μ₂-1)[1/-R -1/∞ ]

-1/90 = (μ₂-1)[1/+30-0]

μ₂-1=13

∴ μ₂ = 43 = 1.33

Hence, the refractive index of the liquid is 1.33.