Wave Optics Test

Result

Wave Optics Test - 24

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the following phenomenon can explain quantum nature of light

A
Photoelectric effect

B
Interference

C
Diffrection

D
Polarisation

Solution

Photoelectric effect explain the quantum nature of light while interference, diffraction and polarization explain the wave nature of light
Question 2
1 / -0

In Young's double experiment, the distance between the slits is reduced to half and the distance between the slit and screen is doubles, then the fringe width

A
Will not change

B
Will become half

C
Will become added

D
Will become four times

Solution

The fringe width is given as:
$$\beta =\dfrac {\lambda D}{d}$$

$$\Rightarrow $$ If $$D$$ becomes twice and $$d$$ becomes half so $$\beta$$ becomes four times.
Question 3
1 / -0

If $$L$$ is the coherent length and $$c$$ the velocity of light, the coherent time is

A
$$cL$$

B
$$\dfrac{L}{c}$$

C
$$\dfrac{c}{L}$$

D
$$\dfrac{1}{Lc}$$

Solution

Coherent time $$=\dfrac{Coherence\ length}{Velocity\ of\ light}=\dfrac{L}{c}$$
Question 4
1 / -0

A star moves away from earth at speed $$0.8\ c$$ while emitting light of frequency $$6\times 10^{14}Hz$$. What frequency will be observed on the earth (in units of $$10\ Hz$$)( $$c=$$ speed of light)

A
$$0.24$$

B
$$1.2$$

C
$$30$$

D
$$3.3$$

Solution

Observed frequency $$c'=c\left(1-\dfrac{v}{c}\right)$$
$$\Rightarrow v'=6\times 10^{14}\left(1-\dfrac{0.8\ c}{c}\right)$$
$$=1.2\times 10^{14}Hz$$
Question 5
1 / -0

The wave front due to a source situated at infinity is :

A
spherical

B
cylindrical

C
planar

D
none of these
Solution
- when you considered it a large distance and measuring justice Mall section of it then it can be considered to be plane wavefront source at Infinity
- example the one coming from sun to earth surface is considered to be plain VU friend from light diverging from a point source will be spherical
- So, the wave front due to a source situated at infinity is planar
- Option C is the right answer
Question 6
1 / -0

In Young's double slit experiment, the two equally bright slits are coherent, but of phase difference $$\dfrac{\pi }{3}$$. If maximum intensity on the screen is $$I_{o}$$, the intensity at the point on the screen equidistant from the slits is:

A
$$I_{o}$$

B
$$\dfrac{I_{o}}{2}$$

C
$$\dfrac{I_{o}}{4}$$

D
$$\dfrac{3I_{o}}{4}$$

Solution

Given$$:\phi =\pi /3$$

$$I=I_{max}cos^{2}(\dfrac{\phi }{2})$$

$$I=I_{0}cos^{2}(\dfrac{\pi /3}{2})$$

$$=I_{0}cos^{2}(\pi /6)$$

$$=I_{0}(\dfrac{\sqrt{3}}{2})^{2} (\because cos\pi /6=\dfrac{\sqrt{3}}{2})$$

$$=I_{0}\dfrac{3}{4}$$

So, intensity on the screen equidistant from the slats is $$\dfrac{3}{4}I_{0}.$$
Question 7
1 / -0

The diameter of the objective of a telescope is $$a$$, its magnifying power is $$m$$ and wavelength of light $$\lambda $$ . The resolving power of the telescope is :

A
$$\dfrac{(1.22\lambda )}{a}$$

B
$$\dfrac{1.22a}{\lambda} $$

C
$$\lambda (1.22a)$$

D
$$\dfrac {a} {1.22\lambda} $$

Solution

Resolving power of telescope:
$$R=\dfrac{1}{ \theta}=\dfrac{a}{1.22 \lambda}$$

where $$\theta$$ is angular resolution, a is diameter of the objective and $$\lambda$$ is wavelength of light.
Question 8
1 / -0

If one of the two slits of a Young's double slit experiment is painted over so that it transmits half the light intensity of the other, then:

A
the fringe system would disappear

B
the bright fringes would be brighter & dark fringes would be darker

C
the dark fringes would be brighter and bright fringes would be darker

D
bright as well as dark fringes would be darker

Solution

Intensity $$I = I_{1} + I_{2} + 2 \sqrt{I_{1}I_{2}} cos \phi$$
$$I_{2} =2 I_{1}$$

Maximum intensity will be at $$\phi = 0 $$ and minimum at $$\phi = \pi/2$$

$$I_{max} = I_{1} + I_{2} + 2I_{1}I_{2} = I_{1} +2 I_{1} +2\sqrt{2} I_{1} = (3+2\sqrt{2}) I_{1} =5.83 I_{1} $$
$$I_{min} = I_{1} + I_{2} + 2I_{1}I_{2} = I_{1} +2 I_{1} -2\sqrt{2} I_{1} = (3-2\sqrt{2}) I_{1}=0.1716 I_{1}$$

Initially when both slits have same intensity then bright fringe intensity is $$4I_{2} = 8 I_{1}$$ and dark fringe is 0

So we see that bright fringe would be bright with reduced intensity and dark fringe would have some intensity now.

$$\therefore$$ Answer C) the dark fringes would be bright and bright fringes would be darker
Question 9
1 / -0

The wave fronts of light wave traveling in vacuum are given by $$x+y+z=c$$ . The angle made by the light ray with the x-axis is:

A
0$$^{\circ}$$

B
45$$^{\circ}$$

C
90$$^{\circ}$$

D
$$cos^{-1}\dfrac{1}{\sqrt{3}}$$

Solution

Normal to the wave front is
(1,1,1)
So, angle between normal and x-axis is angle between light ray and x-axis
So, $$(i+j+k).(i)=\sqrt{3}(1)cos\theta $$
$$cos\theta =\dfrac{1}{\sqrt{3}}$$
$$\theta =cos^{-1}(1/\sqrt{3})$$
Question 10
1 / -0

In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is:

A
$$2:1$$

B
$$4:1$$

C
$$9:1$$

D
$$8:1$$

Solution

Let intensity from the smaller slit be $$I$$ so, the intensity from the bigger slit will be $$4I$$ since it is 4 times of smaller slit.
Now, $$I_{max}=4I+I+2\sqrt{I(4I)}$$
$$=5I+2\sqrt{4I^{2}}$$
$$=5I+2(2I)$$
$$=9I$$
$$Imin=4I+I-2\sqrt{I(4I)}$$
$$=5I-2(2I)$$
$$=I.$$
So, $$\dfrac{I_{max}}{I_{min}}=\dfrac{9I}{I}=\dfrac{9}{1}.$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Wave Optics Test - 24

Result

Wave Optics Test - 24

Score

-

Rank

-

SHARING IS CARING

Question 1

Photoelectric effect

Interference

Diffrection

Polarisation

Solution

Question 2

Will not change

Will become half

Will become added

Will become four times

Solution

Question 3

$$cL$$

$$\dfrac{L}{c}$$

$$\dfrac{c}{L}$$

$$\dfrac{1}{Lc}$$

Solution

Question 4

$$0.24$$

$$1.2$$

$$30$$

$$3.3$$

Solution

Question 5

spherical

cylindrical

planar

none of these

Solution

Question 6

$$I_{o}$$

$$\dfrac{I_{o}}{2}$$

$$\dfrac{I_{o}}{4}$$

$$\dfrac{3I_{o}}{4}$$

Solution

Question 7

$$\dfrac{(1.22\lambda )}{a}$$

$$\dfrac{1.22a}{\lambda} $$

$$\lambda (1.22a)$$

$$\dfrac {a} {1.22\lambda} $$

Solution

Question 8

the fringe system would disappear

the bright fringes would be brighter & dark fringes would be darker

the dark fringes would be brighter and bright fringes would be darker

bright as well as dark fringes would be darker

Solution

Question 9

0$$^{\circ}$$

45$$^{\circ}$$

90$$^{\circ}$$

$$cos^{-1}\dfrac{1}{\sqrt{3}}$$

Solution

Question 10

$$2:1$$

$$4:1$$

$$9:1$$

$$8:1$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.