Ray Optics and Optical Test

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Ray Optics and Optical Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of real depth to apparent depth is called the the

A
refractive index

B
lateral displacement

C
relative density

D
none of these

Solution

refractive index $$ = \dfrac{\text{actual depth}}{\text
{apparent depth}}$$

Option A is correct.
Question 2
1 / -0

If the focal length of a magnifying glass is 2.5 cm, it can magnify an object:

A
15 times

B
11 times

C
2.5 times

D
None of these

Solution

magnifying power MP$$=(0.25m)\phi+1$$ where $$\phi=\dfrac{1}{f}=\dfrac{1}{0.025}=40D$$ is optical power in dioptre.

$$MP=0.25\times40+1=11times$$
Question 3
1 / -0

A number of thin prism of prism angle $$A$$ and refractive index $$\mu$$ are arranged on periphery of circle such that any light ray entering from one prism move along a regular polygon

A
number of prism used will be $$\cfrac { 4\pi }{ (\mu -1)A } $$

B
If $$A$$ is rational, $$\mu$$ must be rational

C
if $$A$$ is rational, $$\mu$$ must be irrational

D
If $$A$$ is irrational, $$\mu$$ must be irrational

Solution

Deviation by one prism: $$\delta = (\mu-1)A$$

$$\therefore$$ Deviation of $$n$$ prism: $$\delta' = n\delta = n(\mu-1)A$$

Deviation of light moving along a regular polygon, $$\delta' = 2\pi$$

$$\therefore$$ $$n(\mu-1)A = 2\pi$$

$$\implies n = \dfrac{2\pi}{(\mu -1)A}$$

Thus if $$A$$ is rational, then $$\mu$$ must be irrational.
Question 4
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A glass prism of refractive index $$1.5$$ is immersed in water $$(\mu=4/3)$$. Light beam incident normally on the face $$AB$$ is totally reflected to reach the face $$BC$$ if

A
$$\sin {\theta}>8/9$$

B
$$2/3< \sin {\theta}< 8/9$$

C
$$\sin {\theta}\le 2/3$$

D
$$\cos {\theta}\ge 8/9$$

Solution

Given -- $$\mu_{g}=1.5, \mu_{w}=4/3$$

As the light beam is normal to prism surface AB, so it will enter in prism without any deviation. Same will happen with water surface. Now the beam will reach to face BC, when it gets total internal reflection from surface AC.

$$\angle\theta'=\angle\theta$$, because angle between two lines is equal to the angle between perpendiculars of the two lines.

We have the condition for total internal reflection,

$$\sin\theta>1/_{w}\mu_{g}$$

or $$\sin\theta>\mu_{w}/\mu_{g}$$

or $$\sin\theta>(4/3)/1.5$$

or $$\sin\theta>8/9$$
Question 5
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Two equilateral prisms are arranged as shown in figure such that a ray of white light is incident at point X The ray of light which emerges out of the prism Q

A
Will show all the colours of spectrum

B
Will be white light

C
Will be orange in colour

D
Will be dull white in colour

Solution

Answer is B.

Optical density is a measure of the refracting power of a medium. In other words, the higher the optical density, the more the light will be refracted or slowed down as it moves through the medium. Optical density is related to refractive index in that materials with a high refractive index will also have a high optical density. Light will travel slower through a medium with a high refractive index and high optical density and faster through a medium which has a low refractive index and a low optical density.
Hence, optical density depends on the velocity of light in that medium.
Question 6
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A student sees the top edge and the bottom centre C of a pool simultaneously from an angle $$\theta $$ above the horizontal as shown in the figure. The refractive index of water which fills up to the top edge of the pool is $$4/3$$. If $$h/x = 7/4$$ then $$cos\theta $$ is:

A
$$\displaystyle \frac { 2 }{ 7 } $$

B
$$\displaystyle \frac { 8 }{ \sqrt [ 3 ]{ 45 } } $$

C
$$\displaystyle \frac { 8 }{ 3\sqrt{ 53 } } $$

D
$$\displaystyle \frac { 8 }{ 21 } $$

Solution

Answer is C.
According to Snell's law,
$$sin(90-\theta )=\mu sinr$$.
$$cos\theta =\dfrac { 4 }{ 3 } \times \dfrac { x/2 }{ \sqrt { { h }^{ 2 }+\dfrac { { x }^{ 2 } }{ 4 } } } $$
$$cos\theta =\dfrac { 2 }{ 3 } \dfrac { 1 }{ \sqrt { \dfrac { { h }^{ 2 } }{ { x }^{ 2 } } +\dfrac { 1 }{ 4 } } } =\dfrac { 2 }{ 3 } \dfrac { 1 }{ \sqrt { \dfrac { 49 }{ 16 } +\dfrac { 1 }{ 4} } } =\dfrac { 2 }{ 3 } \dfrac { 4 }{ \sqrt { 53 } } =\dfrac { 8 }{ 3\sqrt { 53 } } $$.
Hence, $$cos\theta =\dfrac { 8 }{ 3\sqrt { 53 } } $$.
Question 7
1 / -0

A ray of light passes through a rectangular slab of thickness $$t$$. The variation of lateral shift($$\delta$$) with angle of incidence (i) is

A

B

C

D

Solution

Lateral shift ($$\delta$$) is given by
$$\delta = \dfrac{t \sin(i-r)}{\cos r}$$
Hence $$\delta \alpha \sin(i-r)$$
i.e. $$\delta$$ is a sin function of i.
Hence curve in (C) is the option.
Question 8
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Four convex lenses A B, C and D have focal lengths of 5 cm, 5 m, 50 cm and 15 cm respectively.Which one would you choose as a magnifying glass?

A
A

B
B

C
C

D
D

Solution

Answer is A.

When the object is located at a location in front of the focal point, the image will always be located somewhere on the same side of the lens as the object. Regardless of exactly where in front of F the object is located, the image will always be located on the object's side of the lens and somewhere further from the lens. The image is located behind the object.
Therefore, a converging lens produced a virtual image when the object is placed in front of the focal point. For such a position, the image is magnified and upright, thus allowing for easier viewing. Hence, the focal length of the lens should be as small as possible.
Hence, the lens A with a focal length of 5 cm will be chosen as a magnifying glass.
Question 9
1 / -0

Which of the following does not represent correct refraction?

A

B

C

D

Solution

According to the law of refraction if a ray of light passes through rare medium to denser medium it bends towards the normal from its path and vice-versa. So, all the cases following the same, but in the option $$B$$, the ray passes through rare medium to denser medium and its bends away from normal.
Hence, option $$B$$ is correct.
Question 10
1 / -0

When viewed vertically a fish appears to be 4 meter below the surface of the lake. If the index of refraction of water is 1.33, then the true dept of the fish is

A
5.32 metres

B
3.32 metres

C
4.32 metres

D
6.32 metres

Solution

Here the apparent depth of fish is $${ d }_{ apparent }=4 m$$
and we know that
$$refractive\quad index\quad =\frac { actual\quad depth }{ apparent\quad depth } \\ \mu =\frac { { d }_{ actual } }{ { d }_{ apparent } } $$
thus by putting the values we get,
$$1.33=\frac { { d }_{ actual } }{ 4 } $$
$${ d }_{ actual }=5.32m$$
the actual depth of fish is 5.32 m

Option A is correct.