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The path difference between two interfering waves at a point on the screen is \lambda /6. The ratio of intensity at this point and that at the central bright fringe will be : (Assume that intensity due to each slit in same)    

Option: 1 0.853

Option: 2 8.53

Option: 3 0.75

Option: 4 7.5

Answers (1)

best_answer

 

Resultant Intensity of two wave -

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

where

I_{1}= Intencity of wave 1

I_{2}= The intensity of wave 2

\theta = Phase difference

 

 

At path difference  \frac{\lambda }{6}, phase difference is\frac{\pi }{3}

I=I_{O}+I_{O}+2I_{O}\cos \frac{\pi}{3}=3I_{O}.   

and   at the central bright fringe 

 I_{max}=4I_{O}

So the required ratio is \frac{I}{I_{max}}=\frac{3I_{O}}{4I_{O}}=0.75

Posted by

seema garhwal

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