91Ó°ÊÓ

• The refractive index of first medium is$n_1=1.40$.
• The refractive index of the thin film is$n_2=1.46$.
• The refractive index of the third medium is $n_1=1.75$.
• The thickness of the layer is $L=210nm$.

Light that is incident normally on thin films is reflected from both the front and back surfaces, causing interference of the reflected light. When constructive interference happens, it produces bright reflected light, and when entirely destructive interference occurs, it produces a dark region.

The interference of the transmitted rays is similar to the interference of the reflection of light. Here in this case, as $n_1<n_2$and $n_2<n_3$the two transmitted rays have ${180}^{∘}$phase angle difference.

Therefore, the condition for constructive interference is,

$\begin{array}{c}2L=\left(m+\frac{1}{2}\right)\frac{{\mathrm{λ}}_{\max }}{n_2}\\ {\mathrm{λ}}_{\max }=\frac{4L{n}_2}{2m+1}\end{array}$

Calculating the wavelength for first few orders number as below.

For $m=0$:

$\begin{array}{c}{\mathrm{λ}}_1=\frac{4\left(210nm\right)\left(1.46\right)}{2\left(0\right)+1}\\ =1226\text{nm}\end{array}$

For $m=1$:

role="math" localid="1663097372852" $\begin{array}{c}{\mathrm{λ}}_2=\frac{4\left(210nm\right)\left(1.46\right)}{2\left(1\right)+1}\\ =409\text{nm}\end{array}$

For $m=2$:

$\begin{array}{c}{\mathrm{λ}}_3=\frac{4\left(210nm\right)\left(1.46\right)}{2\left(2\right)+1}\\ =245\text{nm}\end{array}$

As $409\text{nm}$lies in visible range, hence the wavelength with maximum intensity of transmitted light is role="math" localid="1663097462853" $409\text{nm}$.