1Didi ans is 0.2 μm
In YDSE both slits produce equal intensities on the screen.A 100% transparent film is placed in front of one of the slits.Now the intensity on the center of screen becomes 75% of previous intensity.The wavelength of light is 6000A and refractive index of glass is 1.5 .Calculate minimum thickness of glass slab.
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4 Answers
1previously,
optical path difference = 0 (at the center f the screen)
after film is kept,
optical path diff = μ X (thickness of slab) = μ t [t = thickness of slab , say]
intial intensity = 4I0 [I0 = intensity of each slit]
later, intensity bcomes 75% i.e. 3I0.
Δθ = Δx X 2pi/λ = 2pi . μt/λ ------------ (1)
intensity = 4I0cos2(Δθ/2) ----------- (2)
from 1 and 2,
cos (Δθ/2) = ± √3/2
=> (Δθ/2) = pi/6 , 5pi/6 , 7pi/6 , 11pi/6 .
we can clearly c, t is min for pi/6
so ... finally ..
t = λ/6μ
1arey nahi nahi main hi wrong hoon... :'(
main phirse isme bhi opt path diff μt ley lii...
yahan (μ-1)t hoga....
then yes the ans is indeed 0.2μm.
shit.