# 6 - Interference and differaction Questions Answers

the ratio of intensities between two coherent sound sources is 4:1. the difference of loudness in decibel between maximum and minimum intensities, when they interfere in space is

(1) 10 log2

(2) 20 log3

(3) 10 log3

(4) 20 log2

**Asked By: HIMANSHU MITTAL**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

I_{1}/I_{2 }= 4/1 so a_{1}/a_{2} = 2/1 so a_{max}/ a_{min} = 2+1/2-1 = 3/1 so I_{max}/ I_{min}= 9/1

now loudness L = 10 log(I/I_{0})

so difference of loudness = L_{1}-L_{2} = 10 log (9I/I) = 10 log9 = 20 log3

two coherent light sources A and B with seperation 2λ are placed the x-axis symmetrically about the origin. they emit light of wavelength λ. obtain the position of maximas on a circle of large radius R lying in the xy plane and with the center at origin

**Asked By: HIMANSHU MITTAL**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

for maxima path diffrence should be integral multiple of λ.

It is given that two coherent light sources A and B with seperation 2λ are placed on the x-axis symmetrically about the origin, and we have to get the points of maxima in the circle so at the point where circle cuts the X axis (path diffrence = 2λ) and where circle cuts the Y axis (path difference = 0) will be the point of maxima. Besides it one point will also be present in the circle in first quadrant where path difference will be λ so it will be a maxima also. it means total 8 points are there

two sound waves having intensities 4 unit and 9 unit coming from two coherent sources arrive at a point along the same line simultaneously in same phase. resultant intensity at the point will be

**Asked By: HIMANSHU MITTAL**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

Use the formule I = I_{1}_{ }+ I_{2} + 2 √I_{1} √I_{2} cosθ

here θ = 0

two sound sources of intensity I and 4I are used in an interference experiment. find the intensity at points where the waves from the two sources are superimposed with a phase difference of

(1) zero

(2) π/2

93)π

**Asked By: HIMANSHU MITTAL**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

Use the formule I = I_{1}_{ }+ I_{2} + 2 √I_{1} √I_{2} cosθ

here θ = 0

two waves are represented by y = Acosωt and y = Asin(ωt - π/6), superimpose on each other. the amplitude of the rsultant wave is

**Asked By: HIMANSHU MITTAL**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

first convert cos into sin then calculate phase difference, its value will be π/2 + π/6 = 2π/3

now use the formula for resultant amplitude

which of the following phenomenon cannot be observed with sound waves?

(1) diffraction

(2) polarisation

(3) reflection

(4) interference

**Asked By: HIMANSHU MITTAL**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

polarisation