Ray Optics and Optical Test

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Ray Optics and Optical Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The dispersive powers of flint glass and crown glass are $$0.053$$ and $$0.034 $$and respectively and their mean refractive indices are $$1.68 $$ and $$1.53$$for white light. Calculate the angle of the flint prism required to form an achromatic combination with a crown glass of refracting angle $$4^0$$

A
$$2^0$$

B
$$4^0$$

C
$$5^0$$

D
$$6^0$$
Question 2
1 / -0

Find the relation between critical angle "C" of a glass prism and its Angle of Prism "A", such that all the rays will reflect back while hitting second surface of prism

A
$$A<2C$$

B
$$2A < C$$

C
$$A>2C$$

D
$$2A>C$$

Solution

Relation between prism angle A & critical angle C such
that ray will always show T I R at BC :
For this $$(r_2) min > C$$ ...(i)
For $$(r_2)$$ min, $$r1$$ should be maximum and
for (r_1)max imax = 90
$$(r_1)max =C $$
$$(r_2)max =AC$$
Now from eq. (i)
$$A C > C $$
$$A > 2C$$
Thus for $$A > 2C$$, all rays are reflected back from the second surface
Question 3
1 / -0

If the tube length of astronomical telescope is $$96$$cm and magnifying power is $$15$$ for normal setting, then the focal length of the objective is _________cm.

A
$$100$$

B
$$90$$

C
$$105$$

D
$$92$$

Solution

$$|m|=\dfrac{f_0}{f_e}=15$$
$$f_0=15f_e$$
Tube length$$=f_0+f_e=96$$
$$15f_e+f_e=96$$
$$f_e=6$$cm
$$f_0=15\times 6=90$$cm.
Question 4
1 / -0

In a hologram, image is stored in the form of

A
Diffraction pattern

B
Photograph

C
Interference pattern

D
Intensity recording

Solution
Question 5
1 / -0

A Point source of Light is placed $$4 \ m$$ below the surface of water of refractive index $$\dfrac{5}{3}$$. the minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is:

A
$$2 \ m $$

B
$$ 6 \ m$$

C
$$4\ m$$

D
$$3\ m$$

Solution

Since the light is in the form of a disc, it implies, that the rays after refraction at the interface of the two media, must be going parallel to the surface.
Applying Snell's law at the interface of the two media, we have
μ1sini=μ2sinr
Given,
μ1=μliquid=53
We know, μ2=μair=1
From the figure,
i=θ and r=90o
⇒53sinθ=1
sinθ=35
From the figure,
sinθ=rr2+h2√
Given, h = 4 cm
rr2+h2√=35
Squaring both sides, we have
r2r2+42=925
r2+16r2=259
16r2=259−1⇒16r2=169
r2=16×916
r=3cm
Diameter, d=2×r=2×3=6cmm
Question 6
1 / -0

Calculate the refractive index of glass with respect to water. It is given that refractive indices of glass and water with respect to air are $$\dfrac {3}{2}$$ and $$\dfrac {4}{3}$$ respectively.

A
$$\dfrac {8}{9}$$

B
$$\dfrac {9}{8}$$

C
$$\dfrac {7}{6}$$

D
$$none\ of\ these$$

Solution

Given:
The refractive index of glass with respect to air is $$n_g=\dfrac{3}{2}$$
The refractive index of water with respect to air is $$n_w=\dfrac{4}{3}$$

Let the refractive index of glass with respect to water is $$n_{gw}$$
According to the definition, $$n_{gw}=\dfrac{n_g}{n_w}=\dfrac{3\times 3}{2\times 4}=\dfrac{9}{8}$$
Question 7
1 / -0

A ray of light is incident on the surface of transparent medium at an angle of $$45^o$$ and is refracted in the medium at an angle of $$30^o$$. What will be the velocity of light in the transparent medium ?

A
$$1.96 \times 10^8 m/s $$

B
$$2.12 \times 10^8 m/s$$

C
$$2.65 \times 10^8 m/s $$

D
$$1.25 \times 10^8 m/s $$

Solution

From Snell's Law, $$\mu = \dfrac{\sin i}{\sin r} = \dfrac{c}{v}$$

$$\Rightarrow v = c\dfrac{\sin r}{\sin i} = 3\times 10^8 \times \dfrac{\sin 30^o}{\sin 45^o}$$

$$\Rightarrow v=3\times 10^8\times \dfrac{\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}} = 2.12 \times 10^8 m/s$$
Question 8
1 / -0

Four rays, 1, 2, 3 and 4 are incident normally on the face PQ of an isosceles prism PQR with apex angles $$\angle Q = 120^{\circ}$$. The refractive indices of the material of the prism for the above rays 1,2, 3 and 4 are 1.85, 1.96, 2.05 and 2.15, respectively and the surrounding medium is air. Then the rays emerging from the face QR are :

A
4 only

B
1 and 2 only

C
3 and 4 only

D
1, 2, 3 and 4

Solution

For critical angle

$$\mu\,sin 30^0=1$$

$$\dfrac{1}{2}=\dfrac{1}{\mu}$$

$$\therefore u=2$$

For $$\mu > 2$$, all rays will suffer TIR from face PR, so all these rays will emerge from face $$QR$$
Question 9
1 / -0

One can not see through fog because:

A
fog absorbed light

B
light is scattered by the droplets in fog

C
light surfers total reflection by the droplets in the fog

D
the refractive index of fog is in infinity

Solution

The fog certains water droplets, When the reflected light from an object travels through fog, it interacts with the water droplets of the fog and is scattered by them. The scattering of light by fog droplets is not uniform in all directions. also it doesn't reaches to the observer eyes so we can not cannot see through fog.
Question 10
1 / -0

In $$n$$ simillar thin prisms of same material and refractive index are arranged in series as shown :

A
if $$n$$ is even number, no net deviation and no net dispersion

B
if $$n$$ is odd number, no net deviation and no net dispersion

C
it depends upon angle of prism

D
no sufficient information

Solution

In given arrangement, if number of prism is even:
Then emerging ray from $$n^{th}$$ prism have no net deviation and no net dispersion. Because of every sets of two prisms diminish their effect.
Hence option A