Wave Optics Test

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Wave Optics Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

In an interference pattern produced by two identical slits, the intensity at the site of the central maximum is I. The intensity at the same spot when either of the two slits is closed is $$I_{\circ}$$. We must have:

A
$$I = I_{\circ}$$

B
$$I = 2I_{\circ}$$

C
$$ I = 4I_{\circ}$$

D
$$ I$$ and $$I_{\circ}$$ are not related

Solution

In case of constructive interference of waves due to 2 slits is, the intensity of central maxima is given by,

$$ E^2 = E_1^2 + E_2^2 + 2 E_1 E_2 $$

where E is the amplitude. For coherent sources, $$ E^2 = I $$

Hence, $$ I = I_0 + I_0 + 2I_o $$

Or $$I = 4$$ $$ I_0 $$
Question 2
1 / -0

Human eye:

A
can detect polarized light

B
cannot detect polarization of light

C
can detect only circularly polarized light

D
can detect only linearly polarized light

Solution

Polarization changes when plane of vibration of polarized light changes.
Human eye is insensitive to change in polarization and hence, cannot detect polarization of light.
Question 3
1 / -0

For the propagation of light wave, medium is required. This is according to

A
Maxwell's theory

B
Huygen's theory

C
Planck's theory

D
Newton's theory

Solution

Huygens suggested that light may be a wave phenomenon produced by mechanical vibrations of an all pervading hypothetical homogenous medium called eather just like those in solids and liguid .This medium was supposed to be mass less with extremely high elasticity and very low density.
Question 4
1 / -0

According to Maxwell , most of the optical properties of light depend on

A
Magnetic vector

B
Electric vector

C
Both Electric and Magnetic vectors

D
Can not be decided

Solution

The experiments on stationary light waves establish that most of the optical properties of light depends on the electric vector, which is also known as light vector.
Question 5
1 / -0

According to Huygens, the ether medium pervading entire universe is

A
Less elastic and more dense

B
Highly elastic and less dense

C
Not elastic

D
Much heavier

Solution

Huygen considered, light needs a medium to propagate called ether which is highly elastic and less denser.
Question 6
1 / -0

In YDSE, the slit widths are in the ratio of $$1: 9$$.The ratio of intensity of the maxima to that of the minima is :

A
$$81 :1$$

B
$$9 : 1$$

C
$$4 : 1$$

D
$$3 : 1$$

Solution

Intensity is proportional to the area of the slit.
As slit widths are in the ratio of $$1 : 9$$
The areas are also in the ratio $$1 : 9$$
Thus Intensities are in the ratio $$1 : 9$$
amplitudes are square root of Intensities
Thus amplitudes are in ratio $$1 : 3$$
Let amplitudes be $$x$$ and $$3x$$
At maxima the amplitudes get added up $$x+3x$$
$$=4x$$
At minima they become $$x-3x=-2x$$
Intensity of maxima to minima is $$\dfrac{16x^{2}}{4x^{2}}=\dfrac{4}{1}$$
Question 7
1 / -0

Two periodic waves of intensities I1 and I2 pass
through a region at the same time in the same
direction. the sum of the maximum and
minimum intensities is -

A
$$ \left ( \sqrt{I_{1}}-\sqrt{I_{2}} \right )^{2}$$

B
$$ 2\left ( I_{1}+I_{2} \right )$$

C
$$ I_{1}+I_{2}$$

D
$$ \left ( \sqrt{I_{1}}+\sqrt{I_{2}} \right )^{2}$$

Solution

$$I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}} \: cos \; \phi$$
$$I_{max}=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}$$
$$I_{min}=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}$$
$$I_{max}+I_{min}=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}+(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}$$
$$=(I_{1}+I_{2}+2\sqrt{I_{1}I_{2}})+(I_{1}+I_{2}-2\sqrt{I_{1}I_{2}})$$
$$=2(I_{1}+I_{2})$$
Question 8
1 / -0

To increase both the resolving power and magnifying power of a telescope

A
Both the focal length and aperture of the objective has to be increased.

B
The focal length of the objective has to be increased.

C
The aperture of the objective has to be increased.

D
The wavelength of light has to be decreased.

Solution

Resolving power, $$R=\dfrac{a}{1.22 \lambda}$$
where, $$a$$ is diameter of objective $$\lambda$$ is wavelength of light
magnifying power $$m=\dfrac{-f_{0}}{f_{e}}\left ( 1+\dfrac{f_{e}}{D} \right )$$
so, decreasing the wavelength of light increases the resolving power and magnifying power of telescope.
Question 9
1 / -0

An electron microscope is superior to an optical microscope in terms of:

A
having better resolving power

B
being easy to handle

C
low cost

D
quickness of observation

Solution

The biggest advantage of an electron microscope over optical microscope is that they have a higher resolution and are therefore capable of a higher magnification ( up to $$2 \ million$$ times ).
However, optical microscopes show a useful magnification up to $$1000-2000 $$ times. This is a limit imposed by the wavelength of light. Electron microscopes, therefore, allow for the visualization of structures that would normally be not visible by optical microscopy.
Question 10
1 / -0

The angular resolution of a telescope of 10 cm diameter at a wavelength of 5000Å is of the order of:

A
10$$^{6}$$ rad

B
$$10^{-2}$$ rad

C
$$10^{-4}$$ rad

D
$$10^{-5}$$ rad

Solution

$$R= \dfrac{1}{\Delta \theta }= \dfrac{a}{1.22\lambda }$$

$$\dfrac{1}{\Delta \theta }= \dfrac{0.10}{1.22\times 5000\times 10^{-10}}$$

$$\Delta \theta = 6.1\times 10^{-6}\ rad$$

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Wave Optics Test - 15

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Wave Optics Test - 15

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SHARING IS CARING

Question 1

$$I = I_{\circ}$$

$$I = 2I_{\circ}$$

$$ I = 4I_{\circ}$$

$$ I$$ and $$I_{\circ}$$ are not related

Solution

Question 2

can detect polarized light

cannot detect polarization of light

can detect only circularly polarized light

can detect only linearly polarized light

Solution

Question 3

Maxwell's theory

Huygen's theory

Planck's theory

Newton's theory

Solution

Question 4

Magnetic vector

Electric vector

Both Electric and Magnetic vectors

Can not be decided

Solution

Question 5

Less elastic and more dense

Highly elastic and less dense

Not elastic

Much heavier

Solution

Question 6

$$81 :1$$

$$9 : 1$$

$$4 : 1$$

$$3 : 1$$

Solution

Question 7

$$ \left ( \sqrt{I_{1}}-\sqrt{I_{2}} \right )^{2}$$

$$ 2\left ( I_{1}+I_{2} \right )$$

$$ I_{1}+I_{2}$$

$$ \left ( \sqrt{I_{1}}+\sqrt{I_{2}} \right )^{2}$$

Solution

Question 8

Both the focal length and aperture of the objective has to be increased.

The focal length of the objective has to be increased.

The aperture of the objective has to be increased.

The wavelength of light has to be decreased.

Solution

Question 9

having better resolving power

being easy to handle

low cost

quickness of observation

Solution

Question 10

10$$^{6}$$ rad

$$10^{-2}$$ rad

$$10^{-4}$$ rad

$$10^{-5}$$ rad

Solution

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