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In YDSE shown in figure a parallel beam ...

In YDSE shown in figure a parallel beam of light is incident on the slits from a medium of refractive index `n_(1)`. The wavelength of light in this medium is `lambda_(1)`. A transparent slab of thickness t and refractive index `n_(3)` is put in front of one slit. The medium between the screen and the plane of the slits is `n_(2)`. The phase difference between the light waves reaching point O (symmetrical, relative to the slits) is

A

`(2 pi)/(n_(1) lambda_(1)) (n_(3) - n_(2)) t`

B

`(2 pi)/(lambda_(1)) (n_(3) - n_(2)) t`

C

`(2 pi n_(1))/(n_(2) lambda_(1)) ((n_(3))/(n_(2)) - 1) t`

D

`(2 pi n_(1))/(lambda_(1)) (n_(3) - n_(2))t`

Text Solution

Verified by Experts

The correct Answer is:
a

Optical path difference the waves `= (n_(3) - n_(2)) t`
`:.` Phase difference `= 2 pi ((n_(3) - n_(2))t)/(lambda_("vacuum")) = 2 pi ((n_(3) - n_(2))t)/(n_(1) lambda_(1))`
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