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Grade: 12th pass
        
Minimum thickness of mica sheet having u=3/2 which should be placed in front of one of the slit in YDSE to reduce the intensity at center of screen to half the maximum intensity is.. . ........ ANS:lambda/2
6 months ago

Answers : (1)

Vikas TU
7263 Points
							
For the refractive index 3/2 mica intoduced,
the path difference becomes,
n*lamda = (u-1)*t
n*lamda = (3/2 – 1)*t
t = 2*n*lamda’
 Now ntensity is inversely proportional to the wavelength.
To reduce the intesity to the half, i.e. for
I = 1/lamda
I/2 = 1/lamda’
lamda’ /lamda = 2
or
lamda’ = lamda/2
 
The wavelngth should become twice of the initial value.
t = lamda/2 would be the thikness for n=1.
6 months ago
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