Wave Optics Test

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Wave Optics Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

A diffraction pattern is obtained using a beam of red light. What happens if the red light is replaced by blue light?

A
No change.

B
Diffraction bands become narrower and crowded together.

C
Band becomes broader and rather apart.

D
Bands disappear altogether.

Solution

We know that the width of the diffraction pattern is given by:
As $$Y= \dfrac{ \lambda D}{d}$$

Since the wavelength of the blue light is smaller than that of the red light, $$\lambda_{blue} < \lambda_{red}$$ and width of diffraction bands being directly proportional to $$\lambda$$, the diffraction bands become narrower and crowded.
Question 2
1 / -0

In Young's double-slit experiment, the angular width of a fringe formed on a distant screen is $$1^o$$. The slit separation is 0.01 mm. The wavelength of the light is

A
0.174 nm

B
$$0.174 \, \overset{0}{A}$$

C
$$0.174 \, \mu m$$

D
$$0.174 \, \times \, 10^{-4} \, m$$

Solution

Angular fringe width, $$\theta \, = \, \dfrac{\lambda}{d}$$
$$\therefore \, \lambda \, = \, \theta d$$
Here, $$d = \, 0.01 mm \, = \, 10^{-5} \, m, \, \theta \, = \, 1^{\circ} \, = \, \dfrac{\pi}{180} \, radians$$
$$\lambda \, = \, \dfrac{\pi}{180} \, \times \, 10^{-5} \, = \, 1.74 \, \times \, 10^{-7} \, m$$
$$= \, 0.174 \, \times \, 10^{-6} \, m \, = \, 0.174 \mu m$$
Question 3
1 / -0

Young's expt. the ratio of intensity at maxima and minima in the interference pattern is The 25 : 9. The ratio of slit width will be

A
4 : 1

B
2 : 1

C
16 : 1

D
8 : 1

Solution

$$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \left( {\dfrac{{25}}{9}} \right) = \left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)$$
$$\dfrac{5}{3} = \left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)$$
$$5{a_1} - 5{a_2} = 3{a_1} + 3{a_2}$$
$$2{a_1} = 8{a_2}$$
$$\dfrac{{{a_1}}}{{{a_2}}} = 4$$
$$\dfrac{{{a_1}}}{{{a_2}}} = \sqrt {\dfrac{{{\omega _1}}}{{{\omega _2}}}} = \dfrac{4}{1}$$
$$\left( {\dfrac{{{\omega _1}}}{{{\omega _2}}}} \right) = \dfrac{{16}}{1}$$
Question 4
1 / -0

Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $$\beta_{G}, \beta_{R}$$ and $$\beta_{B}$$, respectively. Then

A
$$\beta_{G} > \beta_{B} > \beta_{R}$$

B
$$\beta_{B} > \beta_{G} > \beta_{R}$$

C
$$\beta_{R} > \beta_{B} > \beta_{G}$$

D
$$\beta_{R} > \beta_{G} > \beta_{B}$$

Solution

$$\beta =\cfrac { \lambda D }{ d } $$ i.e., $$\beta \propto \lambda $$
and we know that.
$${ \lambda }_{ R }>{ \lambda }_{ G }>{ \lambda }_{ B }$$
So,
$$ { \beta }_{ R }>{ \beta }_{ G }>{ \beta }_{ B }$$
Question 5
1 / -0

For minima to take place between two monochromatic light waves of wavelength $$\lambda ,$$ the path difference should be

A
$$n\lambda $$

B
$$\left( {2n - 1} \right)\dfrac{\lambda }{4}$$

C
$$\left( {2n - 1} \right)\dfrac{\lambda }{2}$$

D
$$\left( {2n - 1} \right)\lambda $$

Solution

For minima,
Path diff $$ = \left( {2n - 1} \right)\frac{\lambda }{2}$$
Question 6
1 / -0

The ratio of resolving power of telescope, when lights of wavelength $$4000\overset{o}{A}$$ and $$5000\overset{o}{A}$$ are used, is _________.

A
$$6 : 5$$

B
$$5 : 4$$

C
$$4 : 5$$

D
$$9 : 1$$

Solution

Resolving power(R.P.) $$\propto \lambda^{-1}$$
Therefore $$\dfrac{R.P._1}{R.P._2}=\dfrac{\lambda_2}{\lambda_1}$$
Given:
$$\lambda_1=4000\overset{o}{A}$$
$$\lambda_2=5000\overset{o}{A}$$
Hence $$\dfrac{R.P._1}{R.P._2}=\dfrac{5000\overset{o}{A}}{4000\overset{o}{A}}$$
$$\dfrac{R.P._1}{R.P._2}=\dfrac{5}{4}$$
Therefore the correct option is (B).
Question 7
1 / -0

In the case of linearly polarized light, the magnitude of the electric field vector

A
is parallel to the direction of propagation

B
does not change with time

C
increases linearly with time

D
varies periodically with time

Solution

In any type of light whether polarised or unpolarised, the magnitude of electric field vector always varies periodically with time.
Actually the change in electric field vector gives rise to periodically changing magnetic field.
Question 8
1 / -0

To increase the magnification of a telescope

A
the objective lens should be of large focal length and eyepiece should be of small focal length.

B
the objective and eyepiece both should be of large focal length.

C
both the objective and eyepiece should be of smaller focal lengths

D
the objective should be of small focal length and eyepiece should be of large focal length

Solution

$$m=\dfrac { { f }_{ o } }{ { f }_{ e } } $$
from the above relation we can see that magnification is directly proportional to the focal length of objective lens.
Question 9
1 / -0

The phenomena which is not explained by Huygen's construction of the wavefront

A
reflection

B
diffraction

C
refraction

D
origin of spectra

Solution

The Huygen's construction of wavefront does not explain the phenomena of origin of spectra.
Question 10
1 / -0

When two waves with same frequency and constant phase difference interfere,

A
there is a gain of energy

B
there is a loss of energy

C
the energy is distributed and the distribution changes with time

D
the energy is redistributed and the distribution remains constant in time

Solution

All the frequency does not change the energy will be redistributed in the regions of constructive interference and distribution remains constant in time.

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Re-Attempt Weekly Quiz Competition

Wave Optics Test - 45

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Wave Optics Test - 45

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SHARING IS CARING

Question 1

No change.

Diffraction bands become narrower and crowded together.

Band becomes broader and rather apart.

Bands disappear altogether.

Solution

Question 2

0.174 nm

$$0.174 \, \overset{0}{A}$$

$$0.174 \, \mu m$$

$$0.174 \, \times \, 10^{-4} \, m$$

Solution

Question 3

4 : 1

2 : 1

16 : 1

8 : 1

Solution

Question 4

$$\beta_{G} > \beta_{B} > \beta_{R}$$

$$\beta_{B} > \beta_{G} > \beta_{R}$$

$$\beta_{R} > \beta_{B} > \beta_{G}$$

$$\beta_{R} > \beta_{G} > \beta_{B}$$

Solution

Question 5

$$n\lambda $$

$$\left( {2n - 1} \right)\dfrac{\lambda }{4}$$

$$\left( {2n - 1} \right)\dfrac{\lambda }{2}$$

$$\left( {2n - 1} \right)\lambda $$

Solution

Question 6

$$6 : 5$$

$$5 : 4$$

$$4 : 5$$

$$9 : 1$$

Solution

Question 7

is parallel to the direction of propagation

does not change with time

increases linearly with time

varies periodically with time

Solution

Question 8

the objective lens should be of large focal length and eyepiece should be of small focal length.

the objective and eyepiece both should be of large focal length.

both the objective and eyepiece should be of smaller focal lengths

the objective should be of small focal length and eyepiece should be of large focal length

Solution

Question 9

reflection

diffraction

refraction

origin of spectra

Solution

Question 10

there is a gain of energy

there is a loss of energy

the energy is distributed and the distribution changes with time

the energy is redistributed and the distribution remains constant in time

Solution

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