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  • Please help! - In a Young's double slit experiment, the path difference , at a certain point on the screen between - Optics - JEE Main

In a Young's double slit experiment, the path difference , at a certain point on the screen between two interfering waves is \frac{1}{8}th of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

 

  • Option 1)

    0.80

  • Option 2)

    0.94

  • Option 3)

    0.85

  • Option 4)

    0.74

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}= Intencity of wave 1

I_{2}= Intencity of wave 2

\theta = Phase difference

\Delta x=\frac{\lambda }{8}

\Delta \phi =\left ( \frac{2\pi }{\lambda } \right )\frac{\lambda }{8}=\frac{\pi }{4}

I=I_{o}\cos ^{2}(\frac{\pi }{8})

\frac{I}{I_{o}}=\cos ^{2}(\frac{\pi }{8})

\approx 0.85

 

 

 


Option 1)

0.80

Option 2)

0.94

Option 3)

0.85

Option 4)

0.74

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