Question 1.

In a Young's Double Slit experiment, the width of one of the two slits is double the other. If the amplitude of light coming from a slit is proportional to the width, find the ratio of the maximum to minimum intensity in the interference pattern.h,

Answer

Let the amplitude of light wave coming from the narrower slit be $a \to$ the amplitude of light wave from the wider slit be 2a

The maximum intensity occurs where the constructive interference takes place and the minimum intensity occurs where the destructive interference takes place

for maximum intensity =2a + a = 3a

for the minimum intensity =2a - a = a

then, the ratio of maximum intensity to minimum intensity will be;

$= \frac{I_{m a x}}{I_{m i n}}$ $= \frac{a_{m a x}^{2}}{a_{m i n}^{2}}$ $= \frac{3^{2} a^{2}}{a^{2}}$ $= 9$

$= 9$

Question 2.

Maximum intensity of the interference pattern in a young double slit experiment is I. Distance between slit is $d = 5 λ$ where $λ$ is wave length of the light source. What will be the intensity of light at a point directly in front of one of the slits.

Answer $(D = 50 λ)$

Path difference $Δ = \frac{y d}{D}$

d = 5 λ

D = 50 λ

y = \frac{d}{2}

= \frac{5}{2}

λ

∴ Δ x = \frac{λ}{4}

Corresponding phase difference

ϕ

ϕ = \frac{2 π}{λ}

Δ x

= \frac{2 π}{λ} \times \frac{λ}{4}

= \frac{π}{2}

\frac{ϕ}{2} = \frac{π}{4}

I_{1} = I \cos^{2} \frac{ϕ}{2}

= I \cos^{2} \frac{π}{4}

= I (\frac{1}{\sqrt{2}})^{2}

= \frac{I}{2}

= \frac{I}{2}

Therefore the intensity of light at a point directly in front of one of the slit is I/2

Question 3.

Two slit in a young's double slit experiment is separated by a distance d. The screen distance is D and it is illuminated by a light of wavelength $λ$ . A point 0 on the screen is $\frac{2 d}{3}$ distance from the point which is directly in front of one of the slits. How far away is 0 from the central bright fringe.

Answer

The central bright fringe is $\frac{d}{2}$

\frac{d}{2}

from the point directly in front of one of the slit.

\frac{d}{2}

∴

distance between central bright fringe and 0

i e \frac{2 d}{3} - \frac{d}{2}

= \frac{d}{6}

distance between central bright fringe and 0

Question 4

= \frac{d}{6}

Two coherent sources of light

S_{1}

and

S_{2}

are placed in a line as shown. Two screens are placed perpendicular to

S_{1} S_{2}

at A and B . It is observed that at point P a bright fringe is seen , then at point Q.

Answer

The path difference at point A due to light from source at

S_{1}

and

S_{2}

S_{2} A - S_{2} A = S_{2} - S_{1}

Since A is a bright point

| S_{2} - S_{1} |

must be an integral multiple of

λ

At B also the path difference is

S_{1} B - S_{2} B = | S_{1} - S_{2} |

Which is an integral multiple of X

∴

B is also a bright point .

Question 5.
In a Young's double slit experiment light of

λ = 5000 A^{\circ}

is incident on two slits. At a point P on the screen the two rays arriving have a path difference of

1 \times 10^{- 6} m

. If the central fringe (zero) is maxima . What is formed at p.

Answer

Path difference

= δ = 1 \times 10^{- 6}

= 2 \times 5 \times 10^{- 7}

= 2 \times 5000 \times 10^{- 10}

= 2 λ

Since path difference is integral multiple

λ

it is a bright fringe

Since

n = 2

Second bright fringe is formed

= \frac{d}{6}

= \frac{d}{6}

= \frac{d}{6}

= \frac{d}{6}

= \frac{d}{6}

= \frac{d}{6}

= \frac{d}{6}

Global education (GE)

Sunday, 23 October 2016

WAVE OPTICS QUESTIONS WITH ANSWERS

Question 1.

In a Young's Double Slit experiment, the width of one of the two slits is double the other. If the amplitude of light coming from a slit is proportional to the width, find the ratio of the maximum to minimum intensity in the interference pattern.h,

Answer

Let the amplitude of light wave coming from the narrower slit be $a \to$ the amplitude of light wave from the wider slit be 2a

The maximum intensity occurs where the constructive interference takes place and the minimum intensity occurs where the destructive interference takes place

for maximum intensity =2a + a = 3a

for the minimum intensity =2a - a = a

then, the ratio of maximum intensity to minimum intensity will be;

$= \frac{I_{m a x}}{I_{m i n}}$ $= \frac{a_{m a x}^{2}}{a_{m i n}^{2}}$ $= \frac{3^{2} a^{2}}{a^{2}}$ $= 9$

$= 9$

$= 9$

Question 2.

Maximum intensity of the interference pattern in a young double slit experiment is I. Distance between slit is $d = 5 λ$ where $λ$ is wave length of the light source. What will be the intensity of light at a point directly in front of one of the slits.

Answer $(D = 50 λ)$

Path difference $Δ = \frac{y d}{D}$

Therefore the intensity of light at a point directly in front of one of the slit is I/2

Question 3.

Answer

The central bright fringe is $\frac{d}{2}$

$λ$

$λ$

$\frac{2 d}{3}$

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Global education (GE)

Sunday, 23 October 2016

WAVE OPTICS QUESTIONS WITH ANSWERS

Question 1.

In a Young's Double Slit experiment, the width of one of the two slits is double the other. If the amplitude of light coming from a slit is proportional to the width, find the ratio of the maximum to minimum intensity in the interference pattern.h,

Answer

Let the amplitude of light wave coming from the narrower slit be a→ the amplitude of light wave from the wider slit be 2a

The maximum intensity occurs where the constructive interference takes place and the minimum intensity occurs where the destructive interference takes place

for maximum intensity =2a + a = 3a

for the minimum intensity =2a - a = a

then, the ratio of maximum intensity to minimum intensity will be;

=ImaxImin=a2maxa2min=32a2a2=9

QQFERGRGTH

Question 2.

Maximum intensity of the interference pattern in a young double slit experiment is I. Distance between slit is d=5λ,where λis wave length of the light source. What will be the intensity of light at a point directly in front of one of the slits.

Answer (D=50λ)

Path difference Δ=ydD

Therefore the intensity of light at a point directly in front of one of the slit is I/2

Question 3.

Answer

The central bright fringe is d2

que

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Let the amplitude of light wave coming from the narrower slit be $a \to$ the amplitude of light wave from the wider slit be 2a

$= \frac{I_{m a x}}{I_{m i n}}$ $= \frac{a_{m a x}^{2}}{a_{m i n}^{2}}$ $= \frac{3^{2} a^{2}}{a^{2}}$ $= 9$

$= 9$

Maximum intensity of the interference pattern in a young double slit experiment is I. Distance between slit is $d = 5 λ$ where $λ$ is wave length of the light source. What will be the intensity of light at a point directly in front of one of the slits.

Answer $(D = 50 λ)$

Path difference $Δ = \frac{y d}{D}$

The central bright fringe is $\frac{d}{2}$

$\frac{2 d}{3}$