Question

White light reflected from a soap film(Refractive Index$=1.5$) has a maxima at $600$nm and a minima at $450$nm with no minimum in between. Then the thickness of the film is _____$\times {10}^{-7}$m.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

Thin film interference in a soap bubble.
For maxima , constructive interference.
$2\mu t\cos r=\frac{(2m+1){\lambda }_1}{2}$
For minima , constructive interference.
$2\mu t\cos r=(m+1){\lambda }_2$
${\lambda }_1$=600 nm ; ${\lambda }_2$=450 nm
t is film thickness and $\mu$ is refractive index of film and is 1.5.
Dividing equation 1 by equation 2
$\frac{(2m+1){\lambda }_1}{2}=(m+1){\lambda }_2$
Putting the values of ${\lambda }_1$ , ${\lambda }_2$
$\frac{(2m+1)600}{2}=(m+1)450$
We get $m=1$.
Putting the value of $m$ in any one of the equations ,
We get for $r=0$,
$t=\frac{(m+1)450}{2\mu }$
$t=\frac{(1+1)450}{2\times 1.5}$

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

Option C is correct.