Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Condition for interference maximum

$\Rightarrow 2\mu t=(n+\frac{1}{2})\lambda ⟶1,n=0,1,2,...$

Condition for interference minimum

$\Rightarrow 2\mu t=n\lambda ⟶2,$

$\left(1\right)\Rightarrow 2\times 1.5\times t=(n+\frac{1}{2})600nm⟶3\left(2\right)\Rightarrow 2\times 1.5\times t=(n+1)\times 450nm⟶4$

Here we took $(n+1)$ instead of $n$ because when we take $n=0$ in first one, we want to take $n=1$ second equation

On solving $3$ and $4$

$t=3\times {10}^{-7}m$