Wave Optics Test

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Wave Optics Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

In YDSE fringe width is $$\beta$$. If distance between the slits is halved and the distance of the screen from the slit is made three times, the new fringe width will be :-

A
$$0.25 \beta$$

B
$$2\beta$$

C
$$6 \beta$$

D
$$8 \beta$$

Solution
Question 2
1 / -0

Light of wavelength 550 nm falls normally on a slit of width 22.0$$\times 10^{-5}$$ cm. The angular potion of the second minimum from the centre maximum will be ( in radians)

A
$$\cfrac{\pi }{13}$$

B
$$\cfrac{\pi }{8}$$

C
$$\cfrac{\pi }{6}$$

D
$$\cfrac{\pi }{4}$$

Solution
Question 3
1 / -0

After reflection from a concave mirror, a plane wave front becomes

A
Cylindrical

B
Spherical

C
Remains planar

D
None of the above

Solution

Given,

As show in figure,

Plane wave front after reflection from concave mirror, turned into spherical wave front.

Hence, spherical wave front will appear.
Question 4
1 / -0

A slit of size $$0.15cm$$ is placed at $$2.1m$$ from a screen. On illuminated it by a light of wavelength $$5 \times {10^{ - 5}}cm$$. The width of diffraction pattern will be:-

A
$$70nm$$

B
$$0.14nm$$

C
$$1.4nm$$

D
$$.014nm$$

Solution
Question 5
1 / -0

The distances of interference point on a screen from two slits are $$18 \, \mu \, m $$ and 12.3 $$\mu$$ m. If the wavelength of light used is $$6 \times 10^{-7}m $$ then the number of dark or bright fringe formed there will be -

A
$$8^{th} $$dark

B
$$9^{th}$$ bright

C
$$10^{th}$$ dark

D
$$11^{th} $$ dark

Solution

$$\begin{array}{l} \Delta =d\sin \theta \\ 5.7\mu m=d\theta \\ \theta =\dfrac { { 5.7\mu n } }{ d } \\ x=\dfrac { { \Delta D } }{ d } \\ \dfrac { \Delta }{ x } =\dfrac { { 5.7 } }{ { 6\times { { 10 }^{ -7 } } } } \\ \dfrac { \Delta }{ x } =\dfrac { { 56 } }{ 6 } =\dfrac { { 19 } }{ 2 } \\ Hence,\, we\, have \\ \dfrac { \Delta }{ x } =9\dfrac { 1 }{ 2 } \end{array}$$
Question 6
1 / -0

The displacement of the interfering sound waves are $${ y }_{ 1 }=4sincot$$ and $${ y }_{ 2 }=3sin\left( cot+\frac { \pi }{ 2 } \right) $$. What is the amplitude of the resultant wave

A
5

B
7

C
1

D
0

Solution
Question 7
1 / -0

Two beams of light will not give rise to an interference pattern, if

A
They are coherent

B
They have the same wavelength

C
They are linearly polarized perpendicular to each other

D
They are not monochromatic

Solution

Two beams having the same wavelength$$,$$ monochromatic or white radiation$$,$$ having the same initial phase $$($$coherent sources$$),$$ can give interference pattern by superposition$$.$$ But their vibrations are perpendicular to each other$$,$$ interference will not be possible$$.$$
Hence,
option $$(C)$$ is correct answer.
Question 8
1 / -0

In an interference pattern of two waves fringe width is $$\beta$$. If the frequency of source is doubled then fringe width will become:-

A
$$\dfrac{1}{2} \beta$$

B
$$ \beta$$

C
$$ 2\beta$$

D
$$\dfrac{3}{2} \beta$$

Solution

Fringe width is given by,
$$\beta=\dfrac{\lambda D}{d}$$
Where, $$\lambda$$ is the wavelength of light, $$D$$ is the distance from slit to screen and $$d$$ is the slit width.
We know,
$$c=\lambda v$$
$$c$$ is speed of light and $$ν$$ is frequency of light.
If $$ν$$ is doubled, then,
$$c=\lambda' 2v=\lambda v$$
$$\implies \lambda'=\dfrac{\lambda}{2}$$
So, the fringe width becomes,
$$\beta'=\dfrac{\lambda' D}{d}=\dfrac 12 \dfrac{\lambda D}{d}=\dfrac{\beta}{2}$$
Thus, the fringe width becomes half of the original value when the frequency of light is doubled.
Question 9
1 / -0

In $$Y.D.S.E$$. If the width of the slits are gradually decreased, then

A
Bright fringe will become brighter and dark fringe become darker

B
Bright fringe become less bright and dark fringe becomes less dark

C
Bright fringe become brighter and dark fringe become lighter

D
Bright fringe become less bright and dark fringe becomes less darker

Solution
Question 10
1 / -0

Which of the following statement is/ are correct?
I. A person suffering from myopia can see distant objects clearly.
II.. The bluish color of water in deep sea is only due to reflection of sky in water.
III. In vacuum, all the colored of the light move with the same speed.
IV. At noon, the sun appears white as light is least scattered

A
I only

B
II only

C
I and II only

D
III and IV only

Solution

Myopia is the inability to see objects clearly when they're too far away. People with myopia can often read or see items within a few feet of their face, but distant objects appear blurred. This condition usually develops in an eyeball that's too long from front to back.
The ocean looks blue because red, orange and yellow (long wavelength light) are absorbed more strongly by water than is blue (short wavelength light). So when white light from the sun enters the ocean, it is mostly the blue that gets returned.