1.

YDSE is carried out in a liquid of refractive index mu=1.3 and a thin film of air is formed in front of the lower slit as shown in the figure. If a maxima of third order is formed at the origin O, find the thickness of the air film. Find the positions of the fourth maxima. The wavelength of light is air is lambda_0 = 0.78 mum and D//d = 1000.

Answer»


Solution :(a) `1-(1/mu)t = ((3lambda)/mu)`
`:. t=((3lambda)/(mu-1))`
` = ((3xx0.78)/(1.3-1))`
`=7.8 mu m`
(b) UPWARDS `(yd)/D-1-(1/mu)t = (4LAMBDA)/mu`
` Solving, we get y=4.2 mm`
Downwards `T1-(1/mu) +(yd)/D=(4lambda)/mu`
Solving, we get `y=06` mm.


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