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In Young's double slit experiment the fr...

In Young's double slit experiment the fringe width is found to be 0.3 mm. Now if a thin glass plate of refracting index 1.5 is placed in the path of any one of light rays coming from the slits then the width of the fringe will be

A

zero

B

0.3 mm

C

0.45 mm

D

0.15 mm

Text Solution

Verified by Experts

The correct Answer is:
B
When a transparent sheet of thickness t and refractive index `mu` is introduced in one of the paths of the interfering waves then the path increases by `(mu-1)`t and the whole pattern shifts by
`y_(0)=(D)/(d)(mu-1)t`
Shifting is towards the side in which the plate is introduced without any change in fringe width as shown in figure.

Therefore, when a thin glass plate of refractive index 1.5 is kept in the path of light from one of the slits, only the fringes get shifted but the fringes width remains unchanged i.e., 0.3 mm.
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