91Ó°ÊÓ

The changes in the direction of propagation of a wave when it meets the boundaries which follow the law of reflection.

The constructive interference in the traveling beam in the film will be multiple of $2\mathrm{Ï€}$.

The thickness is d and the wavelength is role="math" localid="1649150515694" $\mathrm{λ}f=\frac{\mathrm{λ}}{n_f}$

It can be written as $∆âˆ\dots =2\mathrm{Ï€}\frac{2\mathrm{d}}{{\mathrm{λ}}_{\mathrm{f}}}=\mathrm{m}·2\mathrm{Ï€}$

Thus, $\frac{2\mathrm{dnf}}{\mathrm{λ}}=\mathrm{m}=\mathrm{λ}=\frac{2{\mathrm{dn}}_{\mathrm{f}}}{\mathrm{m}}$

The m integer is missing then it is ${\mathrm{λ}}_{min}=\frac{2·5·{10}^{-7}·1.42}{m_{min}}=4·{10}^{-7}⇒m_{min}=3.55$

Similarly ${\mathrm{λ}}_{max}=\frac{2·5·{10}^{-7}·1.42}{m_{max}}=8·{10}^{-7}⇒m_{max}=1.775$

Thus, when the values are m=2 and m=3 the wavelengths will be role="math" localid="1649150962999" $\boxed{{\mathrm{λ}}_1=710nm,{\mathrm{λ}}_2=473nm}$

The destructive interference is m=2.5 and m=3.5 Thus, the interference will be $\boxed{{\mathrm{λ}}_1=568nm,{\mathrm{λ}}_2=406nm}$

Once the light goes back to the air the length becomes the similar.