Wave OpticsHard
Question
Minimum thickness of a mica sheet having μ = 
which should be placed front of one of the slits in YDSE is required to reduce the intensity at the centre of screen to half of maximum intensity is -
which should be placed front of one of the slits in YDSE is required to reduce the intensity at the centre of screen to half of maximum intensity is -Options
A.λ / 4
B.λ / 8
C.λ / 2
D.λ / 3
Solution
Let t be the thickness so corresponding
ᐃx = μt =
t. Also Imax = 4I so I′ = 2I
We know IR = I1 + I2 +
cos 
⇒ I = I + I + 2√I2 cos
⇒ cos 
= 0
⇒ cos
= cos 
or cos 
or 
⇒
⇒ t = 
; 
⇒ t = 
and
⇒ t = 
ᐃx = μt =
t. Also Imax = 4I so I′ = 2IWe know IR = I1 + I2 +
cos ⇒ I = I + I + 2√I2 cos
⇒ cos = 0⇒ cos
= cos or cos or ⇒
⇒ t = ; ⇒ t = and
⇒ t = Create a free account to view solution
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