Question

If wavelengths $4500\mathring{A}$ and $6000\mathring{A}$ are found to be missing in the reflected spectrum in thin air film interference, the thickness of the film for normal incidence is nearly

• A. $9000\mathring{A}$
• B. $10500\mathring{A}$
• C. $5250\mathring{A}$
• D. $4240\mathring{A}$

Answer 1

The correct option is A $9000\mathring{A}$

For missing wavelength, using condition of minima
$2\mu t-\frac{\lambda }{2}=(2n+1)\frac{\lambda }{2}$
$2\mu t=(2n+1)\frac{\lambda }{2}+\frac{\lambda }{2}=(n+1)\lambda$
For air $\mu =1,t=(n+1)\frac{\lambda }{2}$
$\lambda =\frac{2t}{n+1}$
$n+1=\frac{2t}{{\lambda }_1}....\left(i\right)$
$(n-1)+1=\frac{2t}{{\lambda }_2}...\left(ii\right)$
Now, from $\left(i\right)-\left(ii\right)$
$1=2t(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$
$=2t(\frac{1}{4500}-\frac{1}{600})=2t\left(\frac{1500}{4500\times 6000}\right)$

$\Rightarrow t=9000\mathring{A}$