The ratio of the width of the two slits in Youngs Double Slit experiment is 9:1 Prove that in the interference pattern ( I max + I min ) / ( I max - I min ) = 5/3.

Question

The ratio of the width of the two slits in Youngs Double Slit experiment is 9:1 Prove that in the interference pattern ( I max + I min ) / ( I max - I min ) = 5/3.

Adarsh · Answer

The two slits are coherent sources. The intensities of the light coming through these slits is directly proportional to the slit widths. So,I 1 /I 2 =9/1 Let I 1 =9I and I 2 =I )]} 2 I 1 +2√(I 2 +I 1 )-[I 2 I 1 +2√(I 2 +I 1 )}/{I 2 I 1 +2√(I 2 +I 1 )+I 2 I 1 I √( +2 2 +I 1 {I (I max +I min )/(I max – I min )= =2(I 1 +I 2 )/4√(I 1 I 2 )=(2 ×10I)/(4×3I) =10/(2 ×3) =5/3

Adarsh · Answer

I am again posting this answer because while uploading the above answer there was an error The two slits are coherent sources. The intensities of the light coming through these slits is directly proportional to the slit widths. So,I 1 /I 2 =9/1 Let I 1 =9I and I 2 =I (I max +I min )/(I max – I min )= {I 1 +I 2 +2√(I 1 I 2 )+I 1 +I 2 -2√(I 1 I 2 )}/{I 1 +I 2 +2√(I 2 +I 1 )-[I 1 +I 2 -I √( I 1 I 2 )]} =2(I 1 +I 2 )/4√(I 1 I 2 )=(2 ×10I)/(4×3I) =10/(2 ×3) =5/3

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The ratio of the width of the two slits in Youngs Double Slit experiment is 9:1 Prove that in the interference pattern ( I max + I min ) / ( I max - I min ) = 5/3.

The ratio of the width of the two slits in Youngs Double Slit experiment is 9:1 Prove that in the interference pattern
( I_max + I_min) / ( I_max - I_min) = 5/3.

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The ratio of the width of the two slits in Youngs Double Slit experiment is 9:1 Prove that in the interference pattern ( I max + I min ) / ( I max - I min ) = 5/3.

The ratio of the width of the two slits in Youngs Double Slit experiment is 9:1 Prove that in the interference pattern ( Imax + Imin ) / ( Imax - Imin ) = 5/3.

2 Answers

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The ratio of the width of the two slits in Youngs Double Slit experiment is 9:1 Prove that in the interference pattern
( I_max + I_min) / ( I_max - I_min) = 5/3.