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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.

Grade:12

2 Answers

Adarsh
733 Points
6 years ago
The two slits are coherent sources. The intensities of the light coming through these slits is directly proportional to the slit widths.
So,I1/I2 =9/1
Let I1=9I and I2=I
)]}2I1+2√(I2+I1)-[I2I1+2√(I2+I1)}/{I2I1+2√(I2+I1)+I2I1I√(+22+I1{I(Imax+Imin)/(Imax – Imin)=
=2(I1+I2)/4√(I1I2)=(2×10I)/(4×3I)
=10/(2×3)=5/3
Adarsh
733 Points
6 years ago
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,I1/I2 =9/1
Let I1=9I and I2=I
(Imax+Imin)/(Imax – Imin)={I1+I2+2√(I1I2)+I1+I2-2√(I1I2)}/{I1+I2+2√(I2+I1)-[I1+I2-I√(I1I2)]}
=2(I1+I2)/4√(I1I2)=(2×10I)/(4×3I)
=10/(2×3)=5/3

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