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Consider an YDSE that has different slit widths. As a result, amplitude of waves from two silts 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:

Option: 1

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


Option: 2

\frac{I_{0}}{3}\sin ^{2}\frac{\phi}{2}


Option: 3

\frac{I_{0}}{9}\left ( 5+4\cos \phi \right )


Option: 4

\frac{I_0}{9}\left ( 5+8\cos \phi \right )


Answers (1)

best_answer

Resultant Intensity of two wave -

I= I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \theta

- wherein

I_{1}= Intensity of wave 1

I_{2}= Intensity of wave 2

\theta = Phase difference

 As amplitudes are A and 2A, so intensities would be in ratio 1:4,

Let us say they are I and 4I respectively. For maximum intensity, their phase difference is equal to 0o 

The maximum intensity of the resultant wave is,

I_{max}=I_{0}=I+4I+2\sqrt{I \times 4I}=9I

I=\frac{I_{0}}{9}

Intensity at any point,

I'= I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \theta= I +4I +2\sqrt{ 4I^2}\cos \theta

I^{'}=5I+4I\cos \phi =\frac{I_{0}}{9}\left ( 5+4\cos \phi \right )

Posted by

himanshu.meshram

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