Ray Optics and Optical Test

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Ray Optics and Optical Test - 28

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Question 1
1 / -0

If the light moving in a straight line bends by small but fixed angle while entering in different medium , it may be a case of :

A
reflection of light

B
refraction of light

C
scattering of light

D
dispersion of light

Solution

When light travels from one medium to another medium it changes direction and gets bend ,this process is called as refraction of light.It also changes the speed of light in the medium. When light travels from rarer medium to denser medium in bend towards the normal and vice- verca.
Hence it is a case of refraction of light.
Question 2
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A ray of light incident normally on the first face of an equilateral prism undergoes total internal reflection at the second face if refractive index $$ \mu $$ is

A
less than $$ \dfrac { \sqrt { 3 } }{ 2 } $$

B
less than $$ \dfrac { 2 }{ \sqrt { 3 } } $$

C
more than $$ \dfrac { \sqrt { 3 } }{ 2 } $$

D
more than $$ \dfrac { 2 }{ \sqrt { 3 } } $$

Solution

As shown in the figure, on normal incidence at first surface, the angle of incidence at second surface is 60°.
In the border case of total internal reflection just happening,
$$ \mu sin 60 ^\circ = 1 \times sin 90 ^ \circ$$
=> $$\mu = 2/\sqrt {3}$$
Also note that if $$\mu$$ is greater than this value then internal reflection always takes place.
Question 3
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A beam of light consisting of red, green, and blue colors is incident on a right-angled prism. The refractive indices of the material of the prism for the above red, green, and blue wavelengths are 1.39, 1.44, and 1.47, respectively. The prism will

A
separate part of the red color from the green and blue color

B
separate part of the blue color from the red and green color

C
separate all the three colors from one another

D
not separate even partially any color from the other two colors

Solution

The prism will separate part of the red color from the green and blue color as the refractive index are different hence they undergo different deviations
hence correct option is $$A$$.
Question 4
1 / -0

A ray of light enters a spherical drop of water of refractive index $$\mu$$ as shown in fig. Select the correct statement(s).

A
Incident rays are partially reflected at point A

B
Incident rays are totally reflected at point A

C
Incident rays are totally transmitted through A

D
None of these

Solution

From Snell's law:
$$\mu \sin\alpha = \sin\phi$$
$$\sin\alpha = \dfrac{1}{\mu} \times \sin\phi$$
$$0 < \sin\phi < 1$$
$$\dfrac{\sin\phi}{\mu} < \dfrac{1}{\mu}$$
$$\alpha $$ is less than critical angle.
So incident rays are totally transmitted through A.
Hence option 'C' is correct.
Question 5
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A fish is looking at a $$1.0m$$ high plant of the edge of the pond. Will the plant appear shorter or taller than its actual height, to the fish

A
taller

B
shorter

C
no change

D
can't say

Solution

When light from the top of the plant strikes air-water surface it refracts and bends towards the normal and strikes the fish. To the fish the refracted ray is the one appearing to travel from the top of the plant and not the incident ray. When the refracted ray is extended backwards along a straight line, such that it was the original incident ray that didn't undergo refraction, it intersects at a point higher than the actual top of the plant. Hence, the plant appears taller.
Question 6
1 / -0

The refractive index of a prism is $$2$$. The prism can have a maximum refracting angle of

A
$$90^{\circ}$$

B
$$60^{\circ}$$

C
$$45^{\circ}$$

D
$$30^{\circ}$$

Solution

Critical angle of the prism such that it does not refract back into the prism from the emerging side $$=\theta_c=sin^{-1}(\dfrac{1}{\mu})=sin^{-1}\dfrac{1}{2}=30^{\circ}$$

The angles made by the ray inside the prism cannot be greater than $$30^{\circ}$$.

$$A=r_1+r_2$$

If $$A>60^{\circ}$$, the ray does not emerge from the prism. So maximum refracting angle can be $$60^{\circ}$$.
Question 7
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A ray of light enters a rectangular glass slab of refractive index $$\sqrt {3}$$ at an angle of incidence $$60^{\circ}$$. It travels a distance of $$5\ cm$$ inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

A
$$5\sqrt 3\ cm$$

B
$$\dfrac {5}{2} \ cm$$

C
$$5\sqrt {\dfrac{3}{2}}\ cm$$

D
$$5\ cm$$

Solution

Using Snell's Law:

$$\dfrac{sin\ 60^o}{sin\ r_1}=\sqrt3$$

$$\implies sin\ r_1=\dfrac{sin\ 60^o}{\sqrt3}$$

$$\implies sin \ r_1=\dfrac{\sqrt 3}{2}\times \dfrac1{\sqrt3}=\dfrac12$$

$$\implies r_1= 30^o$$

Now, again Applying Snell's Law :

$$sin(i_1- r_1)=\dfrac d5$$

$$\implies d= 5\ sin(i_1- r_1)$$

$$\implies d= 5\ sin(60^o- 30^o)$$

$$\implies d= 5 sin\ 30^o= \dfrac52\ cm$$
Question 8
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A light ray hits the pole of a thin biconvex lens as shown in fig. The angle made by the emergent ray with the optic axis will be approximately

A
$$0^o$$

B
$$(1/3)^o$$

C
$$(2/3)^o$$

D
$$2^o$$

Solution

The curvature of lens becomes perpendicular to the optic axis at the pole. This means that for the snell's law to be applied at that point, $$2^{\circ}$$ is the angle of incidence.

Hence, $$\mu \sin 2^{\circ}=2\mu \sin r_{lens}$$

Again applying Snell's law at the point of emergence,

$$2\mu \sin r_{lens}=3\mu \sin e$$

Hence, $$\sin e=\dfrac{1}{3} \sin 2^{\circ}$$

For small x, $$\sin x \approx x$$

Hence, $$e=(2/3)^{\circ}$$
Question 9
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A given ray of light suffers minimum deviation in an equilateral prism $$P$$. Additional prism $$Q$$ and $$R$$ of identical shape and of the same material as $$P$$ are now added as shown in the figure. The ray will now suffer.

A
Greater deviation

B
No deviation

C
Same deviation as before

D
Total internal reflection

Solution

the deviation due to second prism is same as that of first one but in opposite direction which nullifies the deviation due t first prism so the deviation left is due to only third prism which same as that of first one . option $$C$$ is correct
Question 10
1 / -0

A coin is placed at the bottom of a beaker containing water (refractive index $$= 4/3$$) to a depth of $$12 cm$$. By what height the coin appears to be raised when seen from vertically above?

A
$$9 \ cm$$

B
$$16 \ cm$$

C
$$3 \ cm$$

D
$$4 \ cm$$

Solution

$$ RI=\dfrac{\text{real depth}}{\text{apparent depth}}$$

Substituting, $$\dfrac{4}{3}=\dfrac{12}{apparent depth}$$

Therefore, $$apparent \ depth=\dfrac{12\times3}{4}=9$$

The height by which it appears to be raised is $$12-9= 3cm$$