Wave Optics Test

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Wave Optics Test - 59

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Question 1
1 / -0

What is not an essential condition for an observable interference pattern to occur between the waves two sources?

A
The frequencies of the two sources must be equal

B
The sources must be coherent

C
The sources must emit waves of equal amplitide

D
The waves from the two sources must overlap

Solution

When waves come together they can interfere constructively or destructively. To set up a stable and clear interference pattern, two conditions must be met:
(1) The sources of waves must be coherent. i.e. constant phase difference.
(2) The wave should be monochromatic. i.e. they should gave same wavelength.
So, amplitude can be different.
Option C is correct.
Question 2
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In a double slit experiment, when a thin film of thickness $$t$$ having refractive index $$\mu$$ is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of $$t$$ is ($$\lambda$$ is the wavelength of the light used) :

A
$$\dfrac{\lambda}{2(\mu - 1)}$$

B
$$\dfrac{\lambda}{(2\mu - 1)}$$

C
$$\dfrac{2\lambda}{(\mu - 1)}$$

D
$$\dfrac{\lambda}{(\mu - 1)}$$

Solution

Shift = $$ \Delta X$$

Width of one fringe = $$\lambda$$

$$\Delta X = (\mu - 1)t = 1\lambda$$

For one maximum shift :

$$\boxed { t = \dfrac{\lambda}{\mu - 1}}$$
Question 3
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A plane wavefront of wavelength $$\lambda$$ is incident on a single slit of width a. The angular width of principal maximum is?

A
$$\dfrac{\lambda}{a}$$

B
$$\dfrac{2\lambda}{a}$$

C
$$\dfrac{a}{\lambda}$$

D
$$\dfrac{a}{2\lambda}$$

Solution

$$d \sin\theta =n \lambda$$
For single slit $$: n=1$$
$$ \Rightarrow \sin\theta \dfrac{\lambda}{d}$$
$$=\dfrac{\lambda}{a}$$.
Question 4
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What is the amplitude of resultant wave, when two waves $$y_1=A_1\sin (\omega t-B_1)$$ and $$y_2=A_2\sin (\omega t-B_2)$$ superimpose ?

A
$$A_1+A_2$$

B
$$|A_1-A_2|$$

C
$$\sqrt{A_1^2+A_2^2+2A_1A_2\cos (B_1-B_2)}$$

D
$$\sqrt{A_1^2+A_2^2+2A_1A_2\cos B_1 B_2}$$

Solution
Question 5
1 / -0

If at the focus of a convex lens of focal length $$6$$ metre,a point source of $$36$$ condela is placed, then the illuminance on a screen placed normally to teh emergent beam of light is:

A
$$0$$ lux

B
$$6$$ lux

C
$$1$$ phot

D
$$6$$ phot

Solution
Question 6
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A ray of light is coming from the source $$S$$. If a thin film of thickness $$t$$ and refractive index $$\mu$$ is placed in its path, the increase in length of optical path is :

A
$$\mu t$$

B
$$\dfrac{\mu}{t}$$

C
$$(\mu -1)t$$

D
$$none\ of \ these$$

Solution
Question 7
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To obtain a good photographic print, an exposure of $$2s$$ at a distance of $$1\ m$$ from a $$75\ cd$$ bulb is done. To obtain an equally satisfactory result, what should be the distance, if time of exposure is $$12\ s$$ from a $$50\ cd$$ bulb?

A
$$1\ m$$

B
$$2\ m$$

C
$$3\ m$$

D
$$4\ m$$

Solution
Question 8
1 / -0

Which of these waves can be polarised ?

A
Sound waves

B
Longitudinal waves on a string

C
Transverse waves on a string

D
Light waves
Solution
- Only transverse waves can be polarised. Longitudinal waves (sound waves in air) cannot be polarised. The reason that only transverse wave can be polarised is that their vibrations can be potentially occur in all directions perpendicular to the direction of travel.
- Polarisation of "light wave" is possible only because they can oscillate in more than one orientation because they are transverse in Nature.
- It has no dependence on wavelength and frequencies
- Hence option D is the right answer
Question 9
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If the width of slit is gradually increased, it will be observed experimentally that :

A
bright fringes become reduced in intensity

B
bright fringes become increased in intensity

C
the intensity of minima is strictly zero

D
the fringes become more distinct
Solution
- If the width of the slits are increased, most light pass through slits will be undiffracted
- Hence their is less defraction, the fringes become narrow and less in number
- And the bright fringes reduce its intensity due to less number of fringes
- Hence the option A is the right answer
Question 10
1 / -0

An astronaut is approaching the moon. He sends out a radio signal of frequency $$5000\ MHz$$ and the frequency of echo is different from that of the original frequency by $$100\ kHz$$. His velocity of approach with respect to the moon is:

A
$$2\ km/s$$

B
$$3\ km/s$$

C
$$4\ km/s$$

D
$$5\ km/s$$

Solution