When a Transparent sheet of refractive index `mu=3/2` is placed near one of the slit in YDSE , the intensity at centre of screen reduces to half of maximum intensity. Then, minimum thickness of sheet should be

Options

A. `lamda//4`
B. `lamda//8`
C. `lamda//2`
D. `lamda//3`

Answer

C

Explanation

Let t be the thickness so corresponding
`Deltax=mut=(3)/(2)t`
also `I_(max)=4I`
so `I'=2I`
We know `I_(R)=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cosphi`
`implies2I=I+I+2sqrt(I^(2))cos((3pit)/(lamda))impliescos((3pit)/(lamda))=0`
`impliescos((3pit)/(lamda))=cos((pi)/(2))`
or `cos((3pi)/(2))` or `(5pi)/(2)`
`implies(3pit)/(lamda)=(pi)/(2)impliest=(lamda)/(6),(3pit)/(lamda)=(3pi)/(2)jimpliest=(lamda)/(2)`
and `(3pit)/(lamda)=(5pi)/(2)impliest=(5lamda)/(6)`