Wave Optics Test

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Wave Optics Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

Light is incident at an angle $$\phi$$ with the normal to a plane containing two slits of separation $$d$$. Select the expression that correctly describes the positions of the interference maxima in terms of the incoming angle $$\phi$$ and outgoing angle $$\theta$$

A
$$\sin { \phi } +\sin { \theta } =\left( m+\cfrac { 1 }{ 2 } \right) \cfrac { \lambda }{ d } $$

B
$$d\sin { \theta }=m\lambda$$

C
$$\sin { \phi } -\sin { \theta } =\left( m+1 \right) \cfrac { \lambda }{ d } $$

D
$$\sin { \phi } +\sin { \theta } =m\cfrac { \lambda }{ d } $$

Solution

Path difference of the light travelling, $$\Delta x=RQ +QS$$
$$RQ=dsin\phi ; QS=dsin\theta $$
For the light to interfere constructively,their path difference $$\Delta x=m \lambda$$
So $$dsin\phi +dsin\theta =m\lambda$$
Thus, $$sin\phi +sin\theta=\dfrac{m\lambda}{d}$$
Question 2
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A thin film of refractive index $$1.5$$ and thickness $$4\times { 10 }^{ -5 }cm$$ illuminated by light normal to the surface. What wavelength within the visible spectrum will be intensified in the reflected beam?

A
$$4800$$ $$\mathring { A } $$

B
$$5800$$ $$\mathring { A } $$

C
$$6000$$ $$\mathring { A } $$

D
$$6800$$ $$\mathring { A } $$

Solution

Here $$n_{1}=n_{3}<n_{2}$$
Therefore, a phase change of $$\pi$$ takes place at point A, but not at B. Hence for a constructive interference to be obtained, the condition is-
$$2d\dfrac{n_{2}}{n_{1}}\cos r=(m-\dfrac{1}{2})\lambda$$
Here since the incidence is normal, $$\cos r=1$$
This makes $$\lambda=\dfrac{2d\dfrac{n_{2}}{n_{1}}}{(m-\dfrac{1}{2})}$$
This wavelength lies in the visible region only for m=3
$$\lambda_{m=3}=4800 A^{\circ}$$
Question 3
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Directions For Questions

Interference of light refers to the modification in the intensity distribution of light in a region due to super-position of light waves. Points where the waves arrive in phase are points of maximum intensity (called maxima), whereas minima is obtained when two waves arrive at a point out of the phase. For a production of the interference pattern on a screen, it is essential that the sources producing interference be coherent sources.

...view full instructions

The interference of light was first demonstrated experimentally by

A
Sir Isaac Newton

B
Michelson

C
Fraunhoffer

D
Thomas Young

Solution

Thomas Young demonstrated the phenomenon of interference in water waves.
In 1801, he presented a famous paper to the Royal Society entitled "On the Theory of Light and Colours" which described various interference phenomena, and in 1803 he performed his famous double-slit experiment.
Question 4
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A plane wave of monochromatic light falls normally on a uniform thin film of oil which covers a glass plate. The wavelength of source can be varied continuously. Complete destructive interference is observed for $$\lambda=5000\mathring { A } $$ and $$\lambda=1000\mathring { A } $$ and for no other wavelength in between. If $$\mu$$ of oil is $$1.3$$ and that of glass is $$1.5$$, the thickness of the film will be

A
$$6.738\times { 10 }^{ -5 }cm$$

B
$$5.7\times { 10 }^{ -5 }cm$$

C
$$4\times { 10 }^{ -5 }cm$$

D
$$2.8\times { 10 }^{ -5 }cm$$

Solution

If the order of $$\lambda_1$$ is $$n$$ for $$\lambda_1>\lambda_2$$, then the order of $$\lambda_2$$ is $$n+1$$
Hence, the path difference between interfering waves $$=2\mu_{oil}t=(n-\dfrac{1}{2})\lambda_1=(n+1-\dfrac{1}{2})\lambda_2$$
Here, $$\lambda_1=7000,\lambda_2=5000$$
$$\implies n=3$$
Thus, $$t=\dfrac{(n-\dfrac{1}{2})\lambda_1}{2\mu_{oil}}$$$$=6.738\times 10^{-5}cm$$
Question 5
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In a double-slit experiment, instead of taking slits of equal width, one slit is made twice as wide as the other. Then in the interference pattern

A
the intensities of both the maxima and the minima increase

B
the intensity of the maxima increases and the minima has zero intensity

C
the intensity of the maxima decreases and that of the minima increases

D
the intensity of the maxima decreases and the minima has zero intensity

Solution

Case 1: When slit are of equal width, if $$ { I }_{ o }$$ is intensity of light then,
$$ { I }_{ max }={ I }_{ o }+{ I }_{ o }+2\sqrt { { I }_{ o }{ I }_{ o } } =4{ I }_{ o }$$
$$ { I }_{ min }={ I }_{ o }+{ I }_{ o }-2\sqrt { { I }_{ o }{ I }_{ o } } =0$$
Case 2: When slit are of unequal width, intensity from one slit would increase to twice then,
$$ { I }_{ newmax }={ 2I }_{ o }+{ I }_{ o }+2\sqrt { { 2I }_{ o }{ I }_{ o } } =5.83{ I }_{ o }$$
$$ { I }_{ newmin }=2{ I }_{ o }+{ I }_{ o }-2\sqrt { { 2I }_{ o }{ I }_{ o } } =0.17{ I }_{ o }$$
Therefore, intensities of both the maxima and the minima increase
Question 6
1 / -0

A person wishes to distinguish between two pillars located at a distance of 11 km. What should be the minimum distance between these pillars (resolving power of normal human eye is 1')?

A
1 m

B
3.2 m

C
0.5 m

D
5 m

Solution

Resolving power is given by the distance between two objects to be distinguished per unit distance of objects from the object distinguishing them.
Hence,$$\theta=\dfrac{d}{D}$$
Hence,$$d=\theta D=\dfrac{1}{60}\times \dfrac{\pi}{180}\times 110000=3.2m$$
Question 7
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Consider an YDSE that has different slits width, as a result, amplitudes of waves from two slits are $$A$$ and $$2A$$, respectively. If $$I_{0}$$ be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is $$\phi$$ is :

A
$$I_{0}\cos^{2}\phi$$

B
$$\dfrac {I_{0}}{3}\sin^{2}\dfrac {\phi}{2}$$

C
$$\dfrac {I_{0}}{9}[5 + 4\cos \phi]$$

D
$$\dfrac {I_{0}}{9}[5 + 8\cos \phi]$$

Solution

As amplitudes are $$A$$ and $$2A$$, so intensities would be in the ratio $$1:4$$, let us say $$I$$ and $$4I$$
$${I}_{max}={I}_{0}=I+4I+2\sqrt {4{i}^{2}}=9I$$ $$\Rightarrow$$ $$I={{I}_{0}}{9}$$
Intensity at any point, $${I}^{'}=I+4I+2\sqrt {4{I}^{2}}\cos{\phi}$$
$$I'=5I+4I\cos {\phi}=\cfrac{{I}_{0}}{9}(5+4\cos {\phi})$$
Question 8
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If one of the two slits of Young's double-slit experiment is painted so that it transmits half the light intensity as the second slit, then

A
fringe system will altogether disappear

B
bright fringes will become brighter and the dark fringes will become darker

C
both dark and bright fringes will become darker

D
dark fringes will become less dark and bright fringes will become less bright.

Solution

as the intensity from second slit decreases the bright fringe becomes darker due to constructive interference and dark fringe becomes bright due to destructive interference
option $$D$$ is correct
Question 9
1 / -0

Light waves travel in vacuum along the y-axis. Which of the following may represent the wavefront?

A
x$$=$$constant

B
y$$=$$constant

C
z$$=$$constant

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

Solution

SYNOPSIS WAVE FRONT:
A continuous locus of all the points which are ; in the same phase or state of vibration, is called wavefront. A Point source of light produces a spherical wavefront. A linear source of light produces a cylindrical wave.

If the wave is traveling in vacuum along the y-axis, the wavefront can be a surface perpendicular to the direction of the wave i.e. x-z plane which is $$y=constant$$.
Question 10
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A monochromatic beam of light falls on Young's double slit experiment apparatus as shown in figure. A thin sheet of glass is inserted .in front of lower slit $$S_{2}$$.
The central bright fringe can be obtained :

A
At $$O$$

B
Above $$O$$

C
Below $$O$$

D
Anywhere depending on angle $$\theta$$, thickness of plate $$t$$, and refractive index of glass $$\mu$$

Solution

Because of beam inclination means light reaches $$ { S }_{ 1 }$$ before it reaches $$ { S }_{ 2 }$$, hence an induced time lag before the slits.It means that, without any glass sheet, the central fringe would be above point O because of this inclination
But due to presence of glass sheet, it can be anywhere depending on angle θ, thickness of plate t, and refractive index of glass μ

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Wave Optics Test - 31

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Wave Optics Test - 31

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SHARING IS CARING

Question 1

$$\sin { \phi } +\sin { \theta } =\left( m+\cfrac { 1 }{ 2 } \right) \cfrac { \lambda }{ d } $$

$$d\sin { \theta }=m\lambda$$

$$\sin { \phi } -\sin { \theta } =\left( m+1 \right) \cfrac { \lambda }{ d } $$

$$\sin { \phi } +\sin { \theta } =m\cfrac { \lambda }{ d } $$

Solution

Question 2

$$4800$$ $$\mathring { A } $$

$$5800$$ $$\mathring { A } $$

$$6000$$ $$\mathring { A } $$

$$6800$$ $$\mathring { A } $$

Solution

Question 3

Sir Isaac Newton

Michelson

Fraunhoffer

Thomas Young

Solution

Question 4

$$6.738\times { 10 }^{ -5 }cm$$

$$5.7\times { 10 }^{ -5 }cm$$

$$4\times { 10 }^{ -5 }cm$$

$$2.8\times { 10 }^{ -5 }cm$$

Solution

Question 5

the intensities of both the maxima and the minima increase

the intensity of the maxima increases and the minima has zero intensity

the intensity of the maxima decreases and that of the minima increases

the intensity of the maxima decreases and the minima has zero intensity

Solution

Question 6

1 m

3.2 m

0.5 m

5 m

Solution

Question 7

$$I_{0}\cos^{2}\phi$$

$$\dfrac {I_{0}}{3}\sin^{2}\dfrac {\phi}{2}$$

$$\dfrac {I_{0}}{9}[5 + 4\cos \phi]$$

$$\dfrac {I_{0}}{9}[5 + 8\cos \phi]$$

Solution

Question 8

fringe system will altogether disappear

bright fringes will become brighter and the dark fringes will become darker

both dark and bright fringes will become darker

dark fringes will become less dark and bright fringes will become less bright.

Solution

Question 9

x$$=$$constant

y$$=$$constant

z$$=$$constant

x+y+z$$=$$constant

Solution

Question 10

At $$O$$

Above $$O$$

Below $$O$$

Anywhere depending on angle $$\theta$$, thickness of plate $$t$$, and refractive index of glass $$\mu$$

Solution

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