Wave Optics Test

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Wave Optics Test - 46

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Question 1
1 / -0

A light of wavelength 6300A shine on a two narrow slits separated by a distance 1.0 mm and illuminates a screen at a distance distance 1.5 m away. When one slit is covered by a thin glass of refractive index 1.8 and other slit by a thin glass plate of refractive index $$\Pi $$, the central maxima shifts by $${ 6 }^{ \circ }$$. Both plates have same thickness of 0.5 mm. The value of refractive index $$\Pi $$ of the plate is

A
1.6

B
1.7

C
1.5

D
1.4

Solution
Question 2
1 / -0

A beam of light of wavelength $$600\ nm$$ from a distant source falls on a single slit $$1.00\ mm$$ wide and the resulting diffraction pattern is observed on a screen $$2m$$ away. The distance between the first dark fringes on either side of the central bright fringe is

A
$$1.2\ cm$$

B
$$1.2\ mm$$

C
$$2.4\ cm$$

D
$$2.4\ mm$$

Solution

$$\textbf{Given}:$$ $$\lambda = 600nm$$, d = 1.00 mm, D = 2 m

$$\textbf{Solution}:$$
For the diffraction at a single slit, the position of minima is given by
$$d sin \theta = n \lambda$$
For small value of $$\theta$$,
$$sin \theta\approx \theta=\dfrac{y}{D}$$
$$\therefore \dfrac{dy}{d}=\lambda$$ or
$$y=\dfrac{D}{d} \lambda$$
Substituting the values ,we have
$$y=\dfrac{2 \times 6 \times 10^{-7}}{1 \times 10^{-3}}=1.2 \times 10^{-3}m=1.2 mm$$
$$\therefore$$ Distance between first minima on either side of central side of central maxima=2y=2.4mm.

$$\textbf{Hence D is the correct option}$$
Question 3
1 / -0

A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen placed perpendicular to the length of slits is

A
hyperbola

B
circle

C
straight line

D
parabola

Solution
Question 4
1 / -0

A point sources $$S$$ emitting light of wavelength $$600\ nm$$ is placed at a very small height $$h$$ above a flat reflecting surface $$AB$$ (see figure). The intensity of the reflected light is $$36\%$$ of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance $$D$$ from it.What is the shape of the interference fringes on the screen?

A
Circular.

B
helical

C
eliptical

D
spiral

Solution

Shape of interference fringes is circular due to the symmetry of circular wave bounding from plane number.
Question 5
1 / -0

Two coherent monochromatic light beams of intensities $$I$$ and $$4I$$ are superposed. The maximum and minimum possible intensities in the resulting beam are:

A
$$5I$$ and $$I$$

B
$$5I$$ and $$3I$$

C
$$9I$$ and $$I$$

D
$$9I$$ and $$3I$$

Solution

The correct answer is option (C).

Hint: Apply the formula of maximum and minimum intensities.

Step 1: Calculate the maximum intensity.
$$I_{max}=(\sqrt {I_1}+\sqrt {I_2})^2$$

$$I_{max}=(\sqrt{I}+\sqrt{4I})^2$$
$$I_{max}=(\sqrt{I}+2\sqrt{I})^2$$

$$I_{max}=(3\sqrt I)^2$$
$$I_{max}=9I$$

Step 2: Calculate the minimum intensity.
$$I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2$$

$$I_{min}=(\sqrt{I}-\sqrt{4I})^2$$
$$I_{min}=(\sqrt{I}-2\sqrt{I})^2$$

$$I_{min}=(-\sqrt{I})^2$$
$$I_{min}=I$$

Hence, the maximum and minimum intensities are: $$9I$$ and $$I$$ respectively.
The correct answer is option (C).
Question 6
1 / -0

n identical waves each of intensity $${ 1 }_{ 0 }$$ interfere with each other. The ratio of maximum intensities if the interference is (i) coherent and (ii) incoherent is:

A
$${ n }^{ 2 }$$

B
$$\frac { 1 }{ n } $$

C
$$\frac { 1 }{ { n }^{ 2 } } $$

D
n

Solution

In case of coherent source
$${ l }_{ max }={ \left( \sqrt { { l }_{ 0 } } +\sqrt { { l }_{ 0 } } +\sqrt { { l }_{ 0 } } +.....upto\quad n\quad times \right) }^{ 2 }={ n }^{ 2 }{ l }_{ 0 }$$
In case of incoherent source,
$${ l }_{ max }=n{ l }_{ o }$$
$$\therefore $$ Ratio=n
Question 7
1 / -0

An observer moves towards a stationary source of sound, with a velocity one fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

A
Zero

B
0.5%

C
5%

D
20%

Solution

Apparent frequency
$${ f }^{ 1 }=f\left( \dfrac { V+V/5 }{ V } \right) =\dfrac { 6f }{ 5 } $$
$$\therefore $$ % increase=20
Question 8
1 / -0

Intensity at center in YDSE is $$I_0$$. If one slit is covered, then intensity at center will be:

A
$$I_0$$

B
$$2I_0$$

C
$$I_0/4$$

D
$$I_0/2$$

Solution

As intensity is directly proportional to the square of amplitude. If amplitude of wave of each source is a then intensity will become $${{(2a)}^{2}}=\,4{{a}^{2}}$$. If one source is closed, then new intensity is $${{I}^{'}}$$given by :

$$ {{I}_{0}}=4{{I}^{'}} $$

$$ I'=\dfrac{{{I}_{0}}}{4} $$
Question 9
1 / -0

A small plane mirror is placed at the center of the spherical screen of radius R. A beam of light is falling on the mirror. If the mirror makes n revolutions per second, the speed of light on the screen after reflection from the mirror will be:

A
$$4\pi nR$$

B
$$2\pi nR$$

C
$$nR/2\pi$$

D
$$nR/4\pi$$

Solution
Question 10
1 / -0

If $$\dfrac{I_{1}}{I_{2}}=\dfrac{9}{1}$$ then $$\dfrac{I_{max}}{I_{min}}=?$$

A
$$100 : 64$$

B
$$64 : 100$$

C
$$4 : 1$$

D
$$1 : 4$$

Solution

$$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =\cfrac { 9 }{ 1 } ={ \left( \cfrac { { a }_{ 1 } }{ { a }_{ 2 } } \right) }^{ 2 }\quad \quad \cfrac { { I }_{ max } }{ { I }_{ min } } ={ \left( \cfrac { { a }_{ 1 }+{ { a }_{ 2 } } }{ { a }_{ 1 }-{ a }_{ 2 } } \right) }^{ 2 }\\ \cfrac { { I }_{ max } }{ { I }_{ min } } ={ \left( \cfrac { \cfrac { { a }_{ 1 } }{ { a }_{ 2 } } +{ 1 } }{ \cfrac { { a }_{ 1 } }{ { a }_{ 2 } } -1 } \right) }^{ 2 }={ \left( \cfrac { 3+1 }{ 3-1 } \right) }^{ 2 }={ 2 }^{ 2 }:1\quad =4:1$$