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Light of wavelengthlambda = 500 nm falls...

Light of wavelength`lambda = 500 nm` falls on two narrow slits placed a distance d = ` 50 xx 10^-4` cm
apart, at an angle `phi= 30^@` relative to the slits as shown in figure. On the lower slit a transparent slab of thickness 0.1 mm and refractive index `3/2` is placed. The interference pattern is observed at a distance D=2m from the slits. Then, calculate

(a) position of the central maxima.
(b) the order of maxima at point C of screen .
(c)how many fringes will pass C, if we remove the transparent slab from the lower slit?

A

`30^(@)` above OB

B

`45^(@)` above OB

C

`45^(@)` above OB

D

`30^(@)` below OB

Text Solution

Verified by Experts

The correct Answer is:
D
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