91Ó°ÊÓ

The angle of incidence at the boundary of material 1 is,${\mathrm{θ}}_1={30}^{∘}$.

We can use the formula of Snell’s law to each boundary to find the index of refraction of material 1and angle${\mathrm{θ}}_2$; if${\mathrm{θ}}_1$is changed to${60}^{∘}$and the index of the refraction of material 3 is2.4.

Formula:

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

According to Snell’s law,

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

Therefore,$n_1=n_2$if ${\mathrm{θ}}_1={\mathrm{θ}}_2$.

Since, ${\mathrm{θ}}_1={30}^{∘}$,${\mathrm{θ}}_2={30}^{∘}$

From figure33-52(b), we can conclude that,

${\mathrm{θ}}_2={30}^{∘}$is at$n_2=1.7$

Therefore,

$n_1=1.7$

Therefore, the index of the refraction of material 1 is$1.7$

We have,

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

For,${\mathrm{θ}}_1={60}^{∘}$ and$n_2=2.4$,

Then we can calculate the angle${\mathrm{θ}}_2$as,

$\begin{array}{c}{\mathrm{θ}}_2={\sin }^{−1}\left(\frac{\left(1.7\right)\sin {60}^{°}}{2.4}\right)\\ {\mathrm{θ}}_2={37.83}^{°}\\ ~{38}^{°}\end{array}$

Therefore, angle${\mathrm{θ}}_2$, if the incident angle changed to${60}^{∘}$and material 2 has$n_2=2.4$is${38}^{°}$.