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

We know for maxima
`2mut cos r=((2n+1)lambda_(1))/2`
for minima `2mut cos r=(n+1)lambda_(2)`
Given here, `lambda_(1)=600nm and lambda_(2)=450nm`
Dividing Eq (i) by (ii) and substituting values of `lambda_(1)` and `lambda_(2)` we get
`((2n+1)lambda_(1))/(2(n+1)lambda_(2))=1`
`90m+90=60n+60`
`30 n =30 rArr n=1`
Using one of the equations we get
In given condition `2mut = (2n+1) lambda/2`
`2xx1.5xxt=(2xx1+1)600/2`
`3t=(3xx600)/2`
`rArr t=300nm =3xx10^(-6)m=3units`