Class 12 - Physics - Ray Optics Optical Instruments

Question1.

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:

Size of the candle, h =2.5cm

Image size = h’

Object distance, u = −27 cm

Radius of curvature of the concave mirror, R = −36 cm

Focal length of the concave mirror, f =(R/2) = (-36/2) =-18cm

By mirror formula,

(1/u) + (1/v) = (1/f)

Or (1/v) = (1/f) – (1/u)

= (1/-18) - (1/-27)

= (-3 +2)/ (54)

= (-1/54)

Therefore v = -54cm

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

The magnification of the image is given as:

m= (h’/h)

=-(v/u)

Therefore, h’ = (-v/u) x h

= - (-54/-27) x 2.5

=-5cm

The (-) ive sign shows the image is real and inverted. If the candle is moved closer to the mirror, t

he screen has to be moved away from the mirror to obtain the image on the screen.

However when the candle is moved to a distance less than 18cm from the mirror, the image will become virtual and cannot be obtained on the screen.

Question2.

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:

Given:

Height of the needle h1 = 4.5cm

Focal length of the convex mirror f = 15cm

Object distance u = -12cm

Image distance = v

Using Mirror formula,

 (1/f) = (1/u) + (1/v)

(1/v) = (1/f) – (1/u)

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

= 4 + (5/60)

(1/v)= (9/60)

v = (60/9)

v=+ 6.7cm (As v is (+) ive, the image is erect and virtual)

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

Image size = m = (h2/h1)

h2 = m x h1 = (5/9) x (4.5) = 2.5cm

Hence, the magnification of the image m= (h2/h1) = - (v/u)

= (-20/3)/ (-12) = (5/9)

The height of the image is 2.5cm. 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:

Given:

Actual depth of the tank (h1) = 12.5cm

Apparent depth of the tank (h2) = 9.4

Refractive index = μ

Value of μ can be calculated as:

μ = (h1/ h2)

= (12.5/9.4) = 1.33

Therefore the refractive index of water is 1.33.

When water is replaced by a liquid whose refractive index μ1 =1.63

The actual depth of the needle remains the same, but the apparent depth changes.

Let new apparent depth of the needle = y

Therefore, μ1 = (h1/y)

Or y = (h1/ μ1)

= (12.5/1.63)

=7.67cm

Distance by which the microscope should be moved up = (9.4 – 7.67) =1.73cm

Question 4.

Figures 9.34(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. 9.34(c)].

Physics_Class12_Ray_Optics_Rarer_To_Denser1

Answer:

From the given figure: 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:

μga = (sin i)/(sin r)

= (sin 600)/ (sin 350)

= (0.8660)/ (0.5736)

μga =1.51    (1)

From 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:

μwa = (sin i)/(sin r)

= (sin 600)/ (sin 470)

= (0.8660)/ (0.7314)

μwa = 1.184    (2)

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

μgw = (μga )/ (μwa )

= (1.51)/ (1.184)

μgw =1.275

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

Physics_Class12_Ray_Optics_Denser_To_Rarer

Angle of incidence, i = 45°

Angle of refraction = r

From Snell’s law, r can be calculated as:

(sin i)/ (sin r) = μgw

(sin 450)/ (sin r) = 1.275

sin r = ((1√2/ )/(1.275))

=0.5546

Therefore r = sin-1 (0.5546)

r =38.680

 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:

Given:

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

Refractive index of water, μ = 1.33

From the figure:

Physics_Class12_Ray_Optics_Bulb_In_Tank

Where, Angle of incidence (i) = Angle of refraction(r) = 90°

Since the bulb is a point source, the emergent light can be considered as a circle of radius,

R = (MP/2) = MO=OP

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

μ = (sin r/sin i)

1.33 = (sin 900)/ (sin i)

Or i = sin -1(1/1.33)

i=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

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2.

Question 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.

Answer:

Given:

Angle of minimum deviation, δm = 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:

µ’= (sin (A + δm)/2)/ (sin (A/2))

= (sin (600 + 400))/ (sin (600/2))

= (sin (500))/ (sin (300))

= 1.532

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

Since the prism is placed in water,

 Let δm ‘= the new angle of minimum deviation for the same prism.

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

μgw = (µ’)/ (µ)

= (sin (A + δm ‘)/ (2)) / (sin (A/2))

(sin (A + δm ‘)/ (2)) = (sin (A/2)) (µ’)/ (µ)

(sin (A + δm ‘)/ (2)) = sin (600) (1.532)/ (1.33)

(sin (A + δm ‘)/ (2)) = 0.5759

(A + δm ‘)/ (2) = sin -1(0.5759)

(A + δm ‘)/ (2) = 35.160

(600 + δm ‘) = 70.320

Therefore, δm ‘= (70.320 - 600)

δm ‘=10.320

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:

Refractive index of glass, μ = 1.55

Focal length of the double-convex lens, f = 20 cm

Radius of curvature of one face of the lens = R1

Radius of curvature of the other face of the lens = R2

Radius of curvature of the double-convex lens = R

Therefore, R1 = R and R2 = -R

The value of R can be calculated from Lens – Maker formula :

(1/f) = (μ – 1) [(1/ R1) – (1/ R2)]

(1/20) = (1.55 -1) [(1/R) + (1/R)]

 (1/20) = 0.55 x (2/R)

Therefore R = 0.55 x 2 x20

=22cm

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

 

Question 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 (b) a concave lens of focal length 16cm?

Answer:

In the given scenario, the object is virtual and the image formed is real.

Given:

Object distance, u = +12 cm

(a) Focal length of the convex lens, f = 20 cm

Image distance = v

According to the lens formula, we have the relation:

(1/v) – (1/u) = (1/f)

(1/v) – (1/12) = (1/20)

(1/v) = (1/20) + (1/12)

= (3+5)/ (60)

= (8/60)

Therefore, v = 7.5cm

Hence, the image is formed 7.5 cm away from the lens, toward its right.

(b) Focal length of the concave lens, f = −16 cm

Image distance = v

According to the lens formula, we have the relation:

(1/v) – (1/u) = (1/f)

(1/v) = (-1/16) + (1/12)

(-3 + 4)/ (48)

= (1/48)

Therefore, v = 48cm

Hence, the image is formed 48 cm away from the lens, toward its right.

Question 9.

An object of size 3.0cm is placed 14cm 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?

Answer:

Given:

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:

(1/v) – (1/u) = (1/f)

(1/v) = - (1/21) – (1/14)

= (-2-3)/ (42)

= (-5/42)

Therefore, v = - (42/5) =- 8.4cm

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:

m = (h2)/ (h1) = (v/u)

Therefore h2 = ((-8.4/-14) x 3)

h2 = 1.8cm

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 30cm in contact with a concave lens of focal length 20cm?

Is the system a converging or a diverging lens? Ignore 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:

(1/f) = (1/f1) + (1/f2)

(1/f) = (1/30) – (1/20)

= (2-3)/ (60)

=-(1/60)

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 2.0 cm and an eyepiece of focal length 6.25 cm 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 (b) at infinity?

What is the magnifying power of the microscope in each case?

Answer:

Given:

Focal length of the objective lens, f1 = 2.0cm

Focal length of the eyepiece, f2 = 6.25cm

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

  • Least distance of distinct vision, d’ = 25cm

∴Image distance for the eyepiece, v2= -25cm

Object distance for the eyepiece = u2

According to the lens formula, we have the relation:

(1/ v2) – (1/ u2) = (1/f2)

(1/ u2) = (1/ v2) - (1/f2)

= (-1/25) – (1/6.25)

= (-1-4)/ (25)

= (-5/25)

Therefore, u2 = -5cm

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

Object distance for the objective lens = u1

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)

Therefore, u1 = - 2.5 cm

Magnitude of the object distance,| u1 | = 2.5 cm

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

m = (v1)/ (| u1 |) (1+ (d/f2))

= (10/2.5) (1+ (25/6.25))

=4(1+4)

=20

Hence, the magnifying power of the microscope is 20.

  • The final image is formed at infinity.

∴Image distance for the eyepiece, v2 = ∞

Object distance for the eyepiece = u2

According to the lens formula,

(1/ v2) – (1/ u2) = (1/f2)

(1/ ∞) - (1/ u2) = (1/6.25)

Therefore, u2 = -6.25cm

Image distance for the objective lens, v1 = (d + u2) = (15 – 6.25) = 8.75cm

Object distance for the objective lens = u1

According to the lens formula,

(1/ v1) – (1/ u1) = (1/f1)

(1/ u1) = (1/ v1) - (1/f1)

= (1/8.75) – (1/2.0)

= (2-8.75)/ (17.5)

Therefore, u1= -(17.5)/(6.75) = -2.59cm

Magnitude of the object distance,| u1 | = 2.59cm

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

m = (v1)/ (| u1 |) (1+ (d’/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 9.0mm from the objective in sharp focus. What is the separation between the two lenses?

Calculate the magnifying power of the microscope,

Answer:

Given:

Focal length of the objective lens, fo = 8 mm = 0.8 cm

Focal length of the eyepiece, fe = 2.5 cm

Object distance for the objective lens, uo = −9.0 mm = −0.9 cm

Least distance of distant vision, d = 25 cm

Image distance for the eyepiece, ve = −d = −25 cm

Object distance for the eyepiece = ue

Using the lens formula, we can get the value of ue:

(1/ ve) – (1/ ue) = (1/fe)

(1/ ue) = (1/ ve) – (1/fe)

= (-1/25) – (1/2.5)

= (-1 -10)/ (25)

= (-11/25)

Therefore, ue = - (25/11) = -2.27cm

To get the value of the image distance for the objective lens (vo) by using the lens formula.

(1/ vo) – (1/ uo) = (1/fo)

(1/ vo) = (1/ uo) + (1/fo)

= (1/0.8) – (1/-0.9)

= (0.9 – 0.8)/ (0.72)

= (0.1)/ (0.72)

Therefore, vo = 7.2cm

The distance between the objective lens and the eyepiece = | ue | + vo

= (2.27 + 7.2)

=9.47cm

The magnifying power of the microscope is calculated as:

m = (vo)/ (| uo |) (1+ (d/fe))

= (7.2)/ (|-0.9|) (1 + (25/2.5))

=8(1+10)

=88

Hence, the magnifying power of the microscope is 88.

Question 13.

A small telescope has an objective lens of focal length 144cm 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?

Answer:

Focal length of the objective lens, fo = 144 cm

Focal length of the eyepiece, fe = 6.0 cm

The magnifying power of the telescope is given as:

m = (fo / fe)

= (144/6)

=24

The separation between the objective lens and the eyepiece is calculated as:

L= (fo + fe)

= (144 + 6) = 150cm

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

 

Question 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?

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

Answer:

Given:

Focal length of objective lens fo =15m

Focal length of eye piece fe = 1.0cm = 0.01m

  • Angular magnification m =( fo / fe)

= (15/0.01)

m=1500

  • Diameter of the moon, d = 3.48 × 106 m

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

Let the diameter of image of the moon formed by objective lens= d

The angle subtended by the diameter of the moon = Angle subtended by the image.

=> (3.48 × 106)/ (3.8 × 108) = (d/15)

=>d = 13.74 x 10-2 m

Therefore, d = 13.74cm

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:

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

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

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

 (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

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

Answer:

  • The mirror formula:

(1/v)+ (1/u) = (1/f)

Or (1/v) = (1/f) – (1/u)   (1)

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

∴ f < 0

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

∴u < 0

As the object lies between f and 2f i.e. (2f) < u< f as both u and f are negative.

Or (1/2f)> (1/u)> (1/f)

Or (-1/2f) < (1/u) < (-1/f)

Or ((1/f)-(1/2f)) < ((1/f) – (1/u)) < 0   (2)

Using equation (1):

(1/2f) < (1/v) <0

Therefore, (1/v) is (-) ive or v is negative

(1/2f) < (1/v)

Or 2f> v

  • -v > -2f

Therefore, the image lies beyond 2f.

 

  • For convex mirror, the focal length (f)>0 and (u)<0

(1/v) + (1/u) = (1/f)

Or (1/v) = (1/f) – (1/u)

Using equation (2), (1/v) < 0

Or v>0

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

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

  • For a convex mirror, the focal length (f) is positive.

∴ f > 0

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

∴ u < 0

For image distance v, we have the mirror formula:

(1/v) + (1/u) = (1/f)

Or (1/v) = (1/f) – (1/u)

But u<0

Or (1/v) > (1/f)

Or v>f

Hence, the image formed is diminished and is located between the focus (f) and the pole.

  • For a concave mirror, the focal length (f) is negative.

∴ f < 0

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

∴u < 0

It is placed between the focus (f) and the pole.

Therefore, f> u>0

Or (1/f) < (1/u) <0

(1/f) – (1/u) < 0

For image distance v, we have the mirror formula:

(1/v) + (1/u) = (1/f)

Or (1/v) = (1/f) – (1/u)

Or (1/v) < 0

=> v > 0

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

For u < 0 and v > 0, we can write:

Magnification, m = (v/u) > 1

Hence, the formed image is enlarged.

Question 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 15cm 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:

Given:

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.

μ = (d/d’)

Therefore, d’ = (d/ μ)

= (15/1.5)

=10 cm

The distance at which the pin appears to be raised = (d’ –d)

= (15-10) = 5cm

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

Question 17.

(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68.

The outer covering of the pipe is made of a material of refractive index 1.44.

What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.

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

Physics_Class12_Ray_Optics_LightPipe

Answer:

  • For total internal reflection:

rdμ = (1/sin i’) = (dμ)/( rμ)

Therefore, sin i’ = (rμ)/ (dμ) = (1.44)/ (1.68) = 0.857

Or i’ = sin-1(0.857) ≈ 590

For total internal reflection i’ ≥ 590

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

Maximum angle of reflection, rmax= (900 –i’) = (900 – 590) = 310

Let, imax = maximum angle of incidence.

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

 μ1 = (sin imax)/ (sin rmax)

(sin imax) = μ1 x (sin rmax)

= (1.68 x sin 310)

=1.68 x 0.5150 = 0.8652

 imax = sin -1 (0.8652) ≈600

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

  • If there is no outer covering then,

Refractive index of the outer pipe, μ2 = Refractive index of air =1

For the angle of incidence i = 90°,

Snell’s law at the air − pipe interface can be written as:

(sin i)/ (sin r) = μ2 = 1.68

Or sin r = (sin 900)/ (1.68) = (1/1.68)

r = sin -1(0.5952)

=39.50

Therefore, i’= (900 – 36.50) = 53.50

             As i' > r, all incident rays will suffer total internal reflection.

Question 18.

Answer the following questions:

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

(b) 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?

(c) A diver under water 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?

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

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

Answer:

  • Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.
  • No .A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.
  • The fisherman will look taller than what he actually is. The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. And as it is given that the diver is viewing the fisherman. This shows that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.
  • The apparent depth of water tank decreases when viewed obliquely compared to the depth when viewed normally. It is due to the refraction of light from the surface of water.
  • The refractive index of diamond (μ =2.42) >> than that of ordinary glass (μ = 1.5). Also sin C = (1/ μ). The critical angle C for diamond is very much smaller than that of ordinary glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.

Question 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?

Answer:

Given:

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

Maximum focal length of the convex lens = fmax

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

fmax =(d/4)

= (3/4)

=0.75m

Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Question 20.

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.

Answer:

Given:

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 = (D2 – d2)/ (4D)

= ((902) – (202))/ (4 x 90)

= (770)/ (36)

=21.39cm

Focal length is related to d and D as:

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

Question 21.

(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?

(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 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Answer:

Given:

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

  • The effective focal length of the combination of 2 lenses :

(1/f) = (1/f1) + (1/f2) – (1/ f1 f2)

= (1/30) + (1/-20)-(8/ (30 x (-20))

=- (1/300)

  • f = (-300) cm
  • When parallel beam of light is incident on the convex lens.

From lens formula;

(1/v1) - (1/u1) = (1/f1

Where u1 =Object distance =∞, Image distance = v1

(1/v1) = (1/30) – (1/∞) = (1/30)

Therefore, v1 = 30cm

Using lens formula for concave lens:

(1/v2) - (1/u2) = (1/f2

Where u2 = Object distance = (30-d) = (30-8) = 22cm

Image distance = v2

=> (1/v2) = (1/-20) + (1/22)

=> v2 = -220cm

The parallel incident beam appears to diverge from a point

= (220 - (d/2)) = (220 – 4) = 216cm from the centre of the combination of the two lenses.

  • When parallel beam of light is incident on concave lens:

A virtual image is formed at the focus of the concave lens

(i.e. v2 = -20cm) .The image will act as a real object for the convex lens.

Applying lens formula to the convex lens, we have:

(1/v1) - (1/u1) = (1/f1

Where, u1 = Object distance = − (20 + d) = − (20 + 8) = −28 cm

Image distance = v1

(1/ v1) = (1/30) + (1/-28)

= (14 -15)/ (420) = (-1/420)

Therefore v2 = -420cm

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

Therefore, the answer depends on which side of the lens system the parallel beam is incident.

The notion of effective focal length does not seem to be useful for this combination.

  • Height of the image h1 = 1.5cm

Object distance from the side of the convex lens, u1 = -40cm

Or | u1 | = 40cm

           Applying lens formula to the convex lens, we have:

             (1/v1) - (1/u1) = (1/f1

             => (1/v1) = (1/f1) + (1/u1)

            = (1/30) + (1/-40) = (1/120)

           =>   v1 = +120cm

         And magnification m1 = (v1)/ (u1) = (+120)/ (-40) = -3

         The image by convex lens is magnified, real and inverted.

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

        According to the lens formula:

        (1/v2)- (1/u2) = (1/f2)

        Where, Object distance = u2 = + (120 − 8) = 112 cm.

        Image distance = v2

        (1/v2) = (1/-20) + (1/112)

        = (-112 + 20)/ (2240)

       = (-92)/ (2240)

       Therefore, 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,

Object size = h1 = 1.5 cm

Size of the image = h2

Therefore, h2 = (0.652 x 1.5) = 0.98cm

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:

Physics_Class12_Ray_Optics_Prism

 Given:

Angle of prism, ∠A = 60°

Refractive index of the prism, µ = 1.524

Incident angle = i1

Refracted angle = r1

Angle of incidence at the face AC = r2

Emergent angle = 90° = e

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

(sin e/sinr2) = μ

sin r2 = (1/ μ) sin 90°

= (1/1.524)

sin r2 =0.6562 ≈ 410

Therefore, r2 = sin -1(0.6562)

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

Therefore, r1 = (A – r2)

= (60 – 41)

=190

 

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

μ = (sin i1/sin r1)

sin i1 = (μ x sin r1)

= (1.524 x sin 190) = 0.4962

Therefore, i1 = 29.450

Hence, the angle of incidence is 29.45°

Question 23.

You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will

(a) deviate a pencil of white light without much dispersion,

(b) disperse (and displace) a pencil of white light without much deviation.

Answer:

  • Place two prisms such that they are beside to each other. Their bases are on the opposite sides of the incident white light, with their faces touching each other.
  • When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism,
  • it will recombine and white light will emerge from the combination of the two prisms.
  • If we increase the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal.
  • This combination will disperse the pencil of white light without much deviation.

 

Question 24.

For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm 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:

Least distance of distinct vision, d = 25 cm

Far point of a normal eye, d’ =∞

Converging power of the cornea, Pc = 40D

Least converging power of the eye-lens, Pe =20D

To see the objects at infinity, the eye uses its least converging power. Now, the least converging power of the eye (both cornea and eye lens) is:

Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60 D

Power of the eye-lens is given as:

P = (1/focal length of the eye lens (f))

f= (1/P)

= (1/60D)

= (100/60)

= (5/3) cm

To focus an object at the near point, on the retina,

Object distance (u) = −d = −25 cm

Focal length of the eye-lens = Distance between the cornea and the retina = Image distance

Hence, image distance, v = (5/3) cm

According to the lens formula,

(1/f’) = (1/v) – (1/u); Where, f’ = Focal length

(3/5) + (1/25)

= (15 +1)/ (25)

= (16/25)

Therefore, f = (25/16) cm = (1/64) m

∴Power of the eye-lens = (64 – 40) = 24 D

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

Question 25.

Does short-sightedness (myopia) or long-sightedness (hyper -metropic) imply necessarily that the eye has partially lost its ability of accommodation?

If not, what might cause these defects of vision?

Answer:

No, a person may have normal ability of accommodation of the eye lens and yet suffer from short-sightedness or long-sightedness.

In fact, short-sightedness occurs when the eyeballs gets elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye lens loses its ability of accommodation, the defect is called presbyopia.

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 separate reading glass of power + 2.0 dioptres. Explain what may have happened.

Answer:

Given:

Power P =-1.0D

Therefore, f = (1/P)

= - (1/1.0)

= - 1m

=-100cm

Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.

During old age, the ability of accommodation is partially lost so the near point of the person recedes.

Here u = -25cm; v = -50cm;

Using, (1/f) = (1/v) – (1/u)

= (-1/50) + (1/25)

= (-1 +2)/ (50)

= (1/50)

Therefore f = 50cm

Or P = (1/f) x 100

= (1/50) x100

=+2 dioptres

Therefore, on wearing spectacles of power +2 dioptres, the image of the object lying at a distance of 25cm is formed at a distance of 50cm.

Question 27.

A person looking at a person wearing a shirt with a pattern comprising 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:

This defect is due to a condition called as astigmatism. Here the person is not able to see the horizontal lines;

it is due to the problem in the refracting system of the eye which includes the cornea and the eye lens. This defect can be corrected by using cylindrical lenses.

Question 28.

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

(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

Answer:

Given:

Focal length of the magnifying glass, f = 5cm

Least distance of distance vision, d = 25cm

  • Using lens formula,

(1/v) – (1/u) = (1/f)

Or (1/u) = (1/f) + (1/v)

= (1/-25) + (-1/5)

= (-6/25)

u = (-25/6) = -4.16cm

Hence a person can read a book at closet distance = -4.16cm

For the object at the farthest distance (u1) and image distance (v1) = ∞

(1/v1) – (1/u1) = (1/f1)

(1/u1) = (1/v1) + (1/f1)

= (1/∞) – (1/5)

= (-1/5)

Or u1 = -5

Hence a person can read a book at closest distance is -5cm.

  • Maximum angular magnification, amax = (d/u) =((25)/(25/6))

= (25 x (6/25)) =6

Minimum angular magnification, amin = (d/u’) = (25/5) =5

 

Question 29.

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?

(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.

Answer:

Given:

Area of each square, A =1mm2

Object distance, u= -9cm

Focal length of the lens, f = 10cm

By lens formula, v = (uf)/ (u+f)

= (-9) (10)/ (-9 +10)

v =-90 cm

  • Magnification m =(v/u)

= (-90)/ (-9)

=10cm

=> The size of the image = (1mm x 10) x (1mm x 10)

 = 100mm2

=1cm2

  • Angular magnification = (D/u) = (25/9) = 2.8
  • The magnification in (a) is not the same as the magnifying power in (b).

The magnification magnitude is (|v/u|) and the magnifying power is (d/|u|).

The two quantities will be equal when the image is formed at the near point (25 cm).

 Question 30.

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

(b) What is the magnification in this case?

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

Explain.

Answer:

  • The image should be formed at least distance of distinct vision in order to see the image distinctly. i.e. v=D

From lens formula, u = (fv)/ (f –v)

= (f D)/ (f-D)

= (10 x (-25))/ (10 x (-25))

=-7.14cm

  • Magnification m = (v/u) = (-25)/(-7.14)

=3.5

  • Magnifying power m1 = ( 1 + (D/f))

= (1 + (25/10))

=3.5

Question 31.

What should be the distance between the object in Exercise 9.30 and the 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?

[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the

angular magnification (or magnifying power) of an instrument.]

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 m = √ (A/A0)

=√ (6.25)/ (1)

=2.5

Also, m = (Image distance) (v)/ (Object distance) (u)

Therefore, v = mu

=2.5u   (1)

Applying lens formula,

(1/v) – (1/u) = (1/f)

=> (1/2.5u) – (1/u) = (1/10)

(1/u) = ((1/2.5) – (1/1))

= (1/u) ((1-2.5)/ (2.5))

Therefore u = - (1.5 x 10)/ (2.5)

=> u = -6cm

And v = 2.5u

= - (2.5 x 6)

=-15cm

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) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass.

In what sense then does a magnifying glass provide angular magnification?

(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens.

Does angular magnification change if the eye is moved back?

(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens.

What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?

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

(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece

but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

Answer:

  • Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps us to see the objects placed closer than the least distance of distinct vision (i.e. 25cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
  • Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.
  • The focal length of a convex lens cannot be decreased by a greater amount. 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.
  • The angular magnification produced by the eyepiece of a compound microscope is [(25/fe) +1]

Where, fe = Focal length of the eyepiece

It can be inferred that if fe is small, then angular magnification of the eyepiece will be large.

The angular magnification of the objective lens of a compound microscope is given as: (1/ (|uo/fo|))

Where, Object distance for the objective lens = uo

Focal length of the objective = fo

The magnification is large when (uo > fo). In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since uo is small, fo will be even smaller. Therefore, (fe and fo) are both small in the given condition.

  • For getting complete field of view and all light (refracted), the position of the eye should be at eye-ring which is at a short distance away from eye-piece.

Question 33.

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?

Answer:

Given:

m = -30 (As final image is inverted)

f0 = -1.25cm

fe = 5cm

Assuming final image is formed at a least distance of distinct vision (D =25cm)

Angular magnification m = (v0/u0) [1+ (D/fe)]

Or -30 = - (v0/u0) (1+ (D/fe))

Or -30 = - (v0/u0) (1+ (25/5))

=> (v0/u0) = 5

Using sign conventions,

v0 = -5 u0

From lens formula,

(1/ v0) – (1/ u0) = (1/ f0)

Or (1/-5 u0) – (1/ u0) = (1/1.25)

Therefore, u0 = -1.5cm

This implies the object should be placed at a distance of 1.5cm in front of objective lens of the microscope.

 Also v0 = -5 u0

=-5(-1.5) = + 7.5cm

By lens formula for the eyepiece,

(1/ ve) – (1/ ue) = (1/ fe)

Or (1/ ue) = (1/ ve) - (1/ fe)

= (1/-25) – (1/5)

=-(6/25)

=> ue = - (25/6)

ue = -4.17cm

Therefore the separation between the eye piece and the objective is:

| v0 |+ | u0|

= (7.5) + (4.17)

=11.67cm

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 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)?

Answer:

Given:

Focal length of the objective lens, f0= 140cm

Focal length of the eyepiece, fe = 5cm

D= 25cm

  • For normal adjustment of the telescope.

Magnifying power m = (fo)/ (fe)

(140/5) = 28

  • When the final image is formed at d, the magnifying power of the telescope is given as:

(fo)/ (fe) [1 + (fe)/d]

= (140/5) [1 + (5/25)] = 28[1+0.2] = (28 x 1.2) = 33.6

Question 35.

(a) For the telescope described in Exercise 9.34 (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 25 cm?

Answer:

Given:

Focal length of the objective lens, fo = 140 cm

Focal length of the eyepiece, fe = 5 cm

  • In normal adjustment, the separation between the objective lens and the eyepiece = (fo + fe) = (140 + 5) = 145cm

  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) / (fo)

= (h2) / (140) rad

Where,

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

(1/30) = (h2)/ (140)

Therefore, h2 = (140/30) = 4.7cm

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

 

  • Image is formed at a distance, D = 25 cm

The magnification of the eyepiece is given by the relation:

m= (1 + (D/fe))

= (1+ (25/5))

= (1+5)

=6

Height of the final image = (m h2) = (6 x 4.7)

=28.2m

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 20mm apart.

If the radius of curvature of the large mirror is 220 mm and the small mirror is 140mm, where will the final image of an object at infinity be?

Physics_Class12_Ray_Optics_CassegrainTelescope

Answer:

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

The focal length of the concave mirror f1 = (220/2) = 110mm

Radius of curvature of the secondary mirror, R2 = 140 mm

Hence, focal length of the secondary mirror, f= (140/2) = 70mm

Rays coming from the object which is at infinity are parallel. After striking the concave mirror, the rays converge after reflection.

The rays start moving towards the focus of concave mirror. But as the mirror is placed before focus, the rays strike the convex mirror.

Convex mirror u = (110 -20) = 90mm

Where distance between the 2 mirrors =20mm

Using Mirror formula for secondary mirror,

(1/f) = (1/ u) + (1/ v)

Therefore, (1/90) + (1/ v) = (1/ (140/2))

(1/90) + (1/ v) = (1/70)

Therefore, (1/ v) = (1/70) – (1/90)

= (9-7)/ (630)

(1/ v) = (2/630)

Therefore v =315mm.

Hence, the final image will be formed 315 mm away from the secondary mirror.

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

Physics_Class12_Ray_Optics_LightIncidentNormally_On_Plane_Mirror

Answer:

Given:

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:

tan2 θ = (d/1.5)

Therefore, d = (1.5 x tan70)

d = 0.184m

d=18.4cm

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

Question 38.

Figure 9.37 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?

Physics_Class12_Ray_Optics_EquiConvexLens

Answer:

Given:

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)

= (1/45) – (1/30) = - (1/90)

Therefore, f2 = -90 cm

 Refractive index of the lens = μ1

Radius of curvature of one surface =R.

Hence, the radius of curvature of the other surface = −R.

R can be calculated using the relation = (1/f1) = (μ1 - 1) ((1/R) + (1/-R))

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

Therefore, R = (30)/ (0.5 x 2) = 30cm

Refractive index of the liquid = μ2

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 μ2 of can be calculated using the relation:

(1/f2) = (μ2 - 1) ((1/-R) + (1/ ∞))

(-1/90) = (μ2 - 1) [(1/30) -0]

2 - 1) = (1/3)

Therefore, μ2 = (4/3) = 1.33

Hence, the refractive index of the liquid is 1.33.

 

 

 

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