Ray Optics and Optical Instruments Test

Result

Ray Optics and Optical Instruments Test - 2

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following is correct for the beam which enters the medium?

A
Travel as a cylindrical beam

B
Diverge

C
Converge

D
Diverge near the axis and converge near the periphery

Solution

Since the refractive index is less at beam boundary, the ray at the edges of the beam move faster compared to the axis of beam. Hence the beam converges.
Question 2
4 / -1

Refraction of light from air to glass and from air to water are shown in figure (i) and figure (ii) below. The value of the angle θ in the case of refraction as shown in figure (iii) will be

A
30^∘

B
35^∘

C
60^∘

D
41^∘

Solution

Using Snells law,

⇒ sinθ = sin35^∘ or
θ = 35^∘
Question 3
4 / -1

A ray of light strikes a transparent rectangular slab of refractive index √2 at an angle of incidence of 45°. The angle between the reflected and refracted rays is:

A
75°

B
90°

C
105°

D
120°

Solution

Applying Snell's law at air-glass surface, we get

From figure,
i + θ + 30^∘ = 180^∘ (∵ i = r = 45^∘)
45^∘ + θ + 30^∘ = 180^∘ or
θ = 180^∘ − 75^∘ = 105^∘
Hence, the angle between reflected and refracted rays is 105^∘
Question 4
4 / -1

A ray of light is incident on a thick slab of glass of thickness t as shown in the figure. The emergent ray is parallel to the incident ray but displaced sideways by a distance d. If the angles are small then d is

A

B

C

D

Solution

Lateral shift,

For small angles
sin(i − r) ≈ i − r; cosr ≈ 1
Question 5
4 / -1

A ray incident at a point at an angle of incidence of 60^∘ enters a glass sphere of refractive index √3 and is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is

A
50^∘

B
60^∘

C
90^∘

D
40^∘

Solution

∴ α = 180^o - (r₂′ + i₂) = 180^o −(30^o + 60^o) = 90^o
Question 6
4 / -1

The apparent depth of a needle lying at the bottom of the tank, which is filled with water of refractive index 1.33 to a height of 12.5cm is measured by a microscope to be 9.4cm. If water is replaced by a liquid of refractive index 1.63 upto the same height. What distance would the microscope have to be moved to focus on the needle again?

A
1.73 cm

B
2.13 cm

C
1.5 cm

D
2.9 cm

Solution

Here, Real depth = 12.5cm and

∴ Apparent depth = 12.5/1.63 = 7.67 cm
Now the microscope will have to shift from its initial position to focus 9.4 cm depth object to focus 7.67 cm depth object.
Shift distance = 9.4−7.67
= 1.73 cm
Question 7
4 / -1

A point luminous object (O) is at a distance h from the front face of a glass slab of width d and of refractive index μ. On the back face of a slab is a reflecting plane mirror. An observer sees the image of an object in the mirror as shown in the figure. Distance of image from front face as seen by observer will be :

A

B
2h + 2d

C
h + d

D

Solution

As shown in figure glass slab will form the image of bottom i.e.y mirror MM′ at a depth (d/μ) from its front face. So the distance of object O from virtual mirror mm′ will be h + (d/μ).
Now as a plane mirror forms image behind the mirror at the same distance as the object is in front of it, the distance of image I from mm will be h + (d/μ) and as the distance of virtual mirror from the front face of slab is (d/μ), the distance of image I from front face as seen by
Question 8
4 / -1

A vessel of depth x is half filled with oil of refractive index μ₁ and the other half is filled with water of refractive index μ₂ The apparent depth of the vessel when viewed from above is

A

B

C

D

Solution

∴ Apparent depth of the vessel when viewed from above is
Question 9
4 / -1

Three immiscible liquids of densities d₁ > d₂ > d₃ and refractive indices μ₁ > μ₂ > μ₃ are put in a beaker. The height of each liquid column is h/3. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

A

B

C

D

Solution

Apparent depth of the dot
Question 10
4 / -1

A tank is filled with water to a height of 15.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 8.5 cm. If water is replaced by a liquid of refractive index 1.94 upto the same height, by what distance would the microscope have to be moved to focus on the needle again ?

A
1.00 cm

B
2.37 cm

C
0.51 cm

D
3.93 cm

Solution

Actual depth of the needle in water
h₁ = 15.5cm
Apparent depth of needle in water h₂ = 8.5cm
For another liquid with μ' = 1.94
Apparent depth, H = 15.5/1.94 = 7.99cm
Here, H is less than h₂. Thus to focus the needle again, the microscope should be moved up by the distance 8.5 − 7.99 = 0.51cm