Wave Optics Test

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Wave Optics Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

The limit of resolution of microscope, if the numerical aperture of microscope is 0.12, and the wavelength of light used is 600 nm, is

A
0.3$$\mu $$m

B
1.2 $$\mu $$m

C
2.5$$\mu $$m

D
3$$\mu $$m

Solution

For a microscope, the limit of resolution is given by,
$$X = \dfrac{\lambda}{2A} $$
where $$ \lambda $$ is the wavelength of light used, and A is the numerical aperture.
Hence, substituting the values, $$X = \dfrac{600}{2 \times 0.12} $$,
which gives, $$X= 2.5 \mu m $$
Question 2
1 / -0

In Young's double slit experiment the intensity of the maxima is $$I$$. If the width of each slit is doubled, the intensity of the maxima will be:

A
$$I/2$$

B
$$2I$$

C
$$4I$$

D
$$I$$

Solution

$$I=I_{max}=4I_{0}$$
Now, $$I_{0}$$ is increased to $$2I_{0}$$
So, $$I_{max}=4(2I_{0})=8I_{0}=2I$$
So, maximum intensity is $$2I$$
Question 3
1 / -0

Light waves travel in vaccum along the y-axis. Then the wave front is:

A
$$y=$$ constant

B
$$x=$$ constant

C
$$z=$$ constant

D
$$x+y+z =$$ constant

Solution

waves front are plane perpendicular to the direction of rays.
so, as light is traveling along y - axis, plane perpendicular to y - axis is the x-z plane with any constant value of y.
so, y $$=$$ constant is the wave front plane
Question 4
1 / -0

If accelerating potential increases from $$20\ KV$$ to $$80\ KV$$ in an electron microscope, its resolving power $$R$$ would change to

A
$$\dfrac{R}{4}$$

B
$$4R$$

C
$$2R$$

D
$$\dfrac{R}{2}$$

Solution

$$\dfrac{1}{2}mv^{2}= eV$$

$$mv= \sqrt{2eVm}$$

And $$\lambda = \dfrac{h}{mV}$$

$$\dfrac{\lambda _{0}}{\lambda _{1}}= \dfrac{\sqrt{2eV_{1}m}}{\sqrt{eV_{2}m}}$$

$$\dfrac{\lambda _{2}}{\lambda _{1}}= \dfrac{1}{2}$$

$$\therefore \lambda _{2}=\dfrac{\lambda _{1}}{2}$$

$$R\ \propto \dfrac{1}{\lambda}$$

so $$R$$ would change to $$2R$$.
Question 5
1 / -0

The magnitude of magnifying power of an astronomical telescope is 5, the focal power of its eyepiece is 10 diopters. The focal power of its objective (in diopters) is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$M=5$$

$$P_{e}=10$$

$$\dfrac{1}{f_{e}}=10 $$

$$f_{e}=\dfrac{1}{10}m=10cm$$

$$M=\dfrac{f_{0}}{f_{e}}=5$$

$$f_{0}=50 cm$$

$$P=2D$$
Question 6
1 / -0

A telescope has an objective lens of 10 cm diameter and is situated at a distance of $$1km$$ for two objects. The minimum distance between these two objects, which can be resolved by the telesope, when the mean wavelength of light is 5000Å is of the order of

A
5 cm

B
0.5 mm

C
5 m

D
5 mm

Solution

Resolution power $$= \dfrac {d\lambda}{D} = \dfrac {1000 \times 5000 \times 10^{-10}}{10 \times 10^{-2}} = 5mm$$
Question 7
1 / -0

The resolution limit of the eye is $$1'$$. At a distance $$x\ km$$ from the eye, two persons stand with a lateral separation of 3 meter. For the two persons to be just resolved by the naked eye, $$x$$ should be:

A
$$10 km$$

B
$$15 km$$

C
$$20 km$$

D
$$30 km$$

Solution

Distance = $$x km = 1000x m$$
Resolution limit $$= 1'$$
Lateral separation $$= 3$$ meter
Angle on eye in minutes for 2 person, $$x$$ km away separated by $$3$$ meter $$=$$ $$ \dfrac{3}{1000x} \times \dfrac{180}{ \pi} \times 60$$
So, $$1 =$$ $$ \dfrac{3}{1000 x} \times \dfrac{180}{\pi} \times 60$$

$$\Rightarrow x = 10.3$$ Km ( approximately $$10$$ Km)

Answer. A) $$10$$ Km
Question 8
1 / -0

Which of the following is correct ?

A
Doppler effect in sound and light is asymmetric.

B
Doppler effect in sound and light is symmetric.

C
Doppler effect in sound is asymmetric and light is symmetric.

D
Doppler effect in sound is symmetric and light is asymmetric.

Solution

Doppler effect in sound is asymmetric. This means the change in frequency depends on whether the source is in motion or observer is in motion even though relative velocities are same in both cases.
Doppler effect in sound is asymmetric because sound is mechanical wave requiring material medium and $$v, v_0, v_s$$ are taken with respect to the medium.
Doppler effect in light is symmetric because light waves are electromagnetic (do not require medium).
Question 9
1 / -0

Two point white dots are $$1\ mm$$ apart on a black paper. They are viewed by eye of pupil diameter $$3\ mm$$. Approximately, what is the maximum distance at which these dots can be resolved by the eye? (Take wavelength of light $$= 500\ nm$$)

A
6 m

B
3 m

C
5 m

D
1 m

Solution

$$R.P. = \dfrac{2 \mu \sin \theta}{1.22 \lambda}$$
$$\dfrac{1}{10^{-3}} = \dfrac{2 \sin \theta}{1.22 \times 5 \times 10^{-7}}$$
$$2 \theta = 6 \times 10^{-4}$$
$$\dfrac{d}{D} = 6 \times 10^{-4}$$
$$\therefore D = \dfrac{3 \times 10^{-3}}{6 \times 10^{-4}} = 5 m$$
Question 10
1 / -0

A person wants to see two pillars from a distance of 11 km, separately. The distance between the pillars must be approximately

A
3.2m

B
1m

C
0.25 m

D
0.5 m

Solution

Resolving power of eye = $$ 1' = \dfrac {\pi} {60 \times 180} radians$$ (approximately)

Let the separation between pillars $$= x m$$
Distance $$= 11000 m$$

=> $$\dfrac {x} {11000} = \dfrac {\pi}{60 \times 180}$$

=> $$ x = 3.2 m$$ (approximately)

Answer. A)$$ 3.2m $$

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Wave Optics Test - 25

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Wave Optics Test - 25

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Question 1

0.3$$\mu $$m

1.2 $$\mu $$m

2.5$$\mu $$m

3$$\mu $$m

Solution

Question 2

$$I/2$$

$$2I$$

$$4I$$

$$I$$

Solution

Question 3

$$y=$$ constant

$$x=$$ constant

$$z=$$ constant

$$x+y+z =$$ constant

Solution

Question 4

$$\dfrac{R}{4}$$

$$4R$$

$$2R$$

$$\dfrac{R}{2}$$

Solution

Question 5

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 6

5 cm

0.5 mm

5 m

5 mm

Solution

Question 7

$$10 km$$

$$15 km$$

$$20 km$$

$$30 km$$

Solution

Question 8

Doppler effect in sound and light is asymmetric.

Doppler effect in sound and light is symmetric.

Doppler effect in sound is asymmetric and light is symmetric.

Doppler effect in sound is symmetric and light is asymmetric.

Solution

Question 9

6 m

3 m

5 m

1 m

Solution

Question 10

3.2m

1m

0.25 m

0.5 m

Solution

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