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 princpal maxima front 'O' is

Options

A. `(D)/(4)`
B. `(D)/(16)`
C. `(D)/(8)`
D. `D`

Answer

B

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.