White light reflected from a soap film (Refractive index `=1.5` ) has a maxima at 600 mm and a minima at 450 nm at with no minimum in between. Then, the thickness of the film is

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

The condition for an interference maximum in the light reflected from an soap film of thickness t is
`2mu t(n + (1)/(2)) " where " n =0,1,2….`
and that for interference minimum is
`2mu t = n lamda` where n=1,2,3,..
Here `mu= 1.5`. Now there is a maximum for `lamda = 600` nm and there is minimum for `lamda = 450nm`
`:. 2 xx 1.5 xx t = (n + (1)/(2)) 600` ...(i)
`2 xx 1.5 xx t =(n+1)450` ..(ii)
In view of Eq.(i), we have taken the integer `(n +1)` rather than n in Eq (ii)
Comparing (i) and (ii), we get
`(n+(1)/(2)) 600 = (n+1) 450 " " :. n=1`
Substituting n=1, in Eq.(i), we get
`2 xx 1.5 xx t = (3)/(2) xx 600nm`
`t= (3 xx 600 xx 10^(-9))/(2 xx 2 xx 1.5) = 3 xx 10^(-7) m`