A small transparent slab `(mu = 1.5)` is placed along `AS_(2)` as shown.

`AC=CO=D,S_(1)C=S_(2)S=d lt ltD`
The distance of first minima above O is

Options

A. `(3D)/(sqrt(43))`
B. `(D)/(sqrt(247))`
C. `(3D)/(sqrt(247))`
D. `(5D)/(sqrt(247))`

Answer

C

Explanation

Path difference `=2d sin theta + (mu-1)l`
`:.` For principal maxima,
`2d sin theta + 0.5l = 0`
`sin theta_(0) = (-l)/(4d) = (-1)/(16)` , `((:' l = (d)/(4))`
`:. OP = D tan theta_(0) ~~ -(D)/(16)`
for the first minima :
`:. 2d sin theta+0.5l = +-(lambda)/(2)`
The first minima the positive side is at distance
`D tan theta^(+) = D (3)/(sqrt(16^(2)-3^(2)))` above O.
In the `-ve` side, the distance will be
`D tan theta^(-) = D (5)/(sqrt(16^(2)-5^(2)))` below O.