91Ó°ÊÓ

The angle of incidence at the boundary of material 1 is ${\text{θ}}_{\text{1}}{\text{=40}}^{\text{0}}$.

Use the formula of Snell’s law to each boundary to find the index of refraction of material 1and angle${\mathrm{θ}}_3$, it${\mathrm{θ}}_1$is changed to${70}^0$and the index of refraction of material 3 is$\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{.}$

The formula is as follows:

${\mathit{n}}_{\mathbf{1}}\mathit{s}\mathit{i}\mathit{n}{\mathit{θ}}_{\mathbf{1}}\mathbf{=}{\mathit{n}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}{\mathit{θ}}_{\mathbf{2}}$

Where,

${\mathit{θ}}_{\mathbf{1}}$= angle of incidence,

${\mathit{θ}}_{\mathbf{2}}$= angle of refraction,

n₁= index of refraction of the incident medium,

n₂= index of refraction of the refractive medium,

$\begin{array}{l}{n}_1{\text{²õ¾±²Ôθ}}_{\text{1}}=n_2{\text{²õ¾±²Ôθ}}_{\text{2}}\\ n_2{\text{²õ¾±²Ôθ}}_{\text{2}}=n_3{\text{²õ¾±²Ôθ}}_{\text{3}}\end{array}$

Therefore,

$n_1{\text{²õ¾±²Ôθ}}_{\text{1}}=n_3{\text{²õ¾±²Ôθ}}_{\text{3}}{\text{θ}}_{\text{1}}={\text{θ}}_{\text{3}}$

So,

$\text{When}{n}_1=n_3$

Since.${\text{θ}}_{\text{1}}{\text{=40}}^{\text{0}}$Then,${\text{θ}}_{\text{3}}{\text{=40}}^{\text{0}}$.

Figure 33.50 (b),${\text{θ}}_{\text{3}}{\text{=40}}^{\text{0}}$is at$n_3=\text{1.6}.$

Therefore,

$n_1=\text{1.6}$

Therefore, the index of refraction of a material 1 is $\text{1.6}$.

For,${\text{θ}}_{\text{1}}{\text{=70}}^{\text{0}}$ and ${\text{n}}_{\text{3}}\text{=2.4}$,

$n_1{\text{²õ¾±²Ôθ}}_{\text{1}}=n_3{\text{²õ¾±²Ôθ}}_{\text{3}}$

$\begin{array}{l}{\text{θ}}_{\text{3}}{\text{=sin}}^{\text{-1}}\left(\frac{n_1{\text{sin70}}^{\text{0}}}{\text{2.4}}\right)\\ {\text{θ}}_{\text{3}}{\text{=39}}^{\text{0}}\end{array}$

Therefore, it${\text{θ}}_{\text{1}}$ is changed to ${\text{70}}^{\text{0}}$and the index of refraction of material 3 is $\text{2.4}$,${\text{θ}}_{\text{3}}$ is${\text{39}}^{\text{0}}$ .

Using Snell’s law, the angle of incidence and the refraction index of the layers of a multilayered material can be found.