Question

A thin film of refractive index $1.5$ and thickness $4\times {10}^{-5}cm$ illuminated by light normal to the surface. What wavelength within the visible spectrum will be intensified in the reflected beam?

• A. $4800$ $\mathring{A}$
• B. $5800$ $\mathring{A}$
• C. $6000$ $\mathring{A}$
• D. $6800$ $\mathring{A}$

Answer 1

The correct option is A $4800$ $\mathring{A}$

Here $n_1=n_3<n_2$

Therefore, a phase change of $\pi$ takes place at point A, but not at B. Hence for a constructive interference to be obtained, the condition is-

$2d\frac{n_2}{n_1}\cos r=(m-\frac{1}{2})\lambda$

Here since the incidence is normal, $\cos r=1$

This makes $\lambda =\frac{2d\frac{n_2}{n_1}}{(m-\frac{1}{2})}$

This wavelength lies in the visible region only for m=3

${\lambda }_{m=3}=4800A^{\circ }$