91Ó°ÊÓ

We are given that the refractive index of air $n_{air}=1.00$and refractive index of glass is $n_{glass}=1.50$. We have to find expression for angle of incidence ${\mathrm{θ}}_1$

By applying Snell's Law $n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

Where,

$n_1=$refractive index of air

$n_2=$refractive index of transparent medium

According to question ${\mathrm{θ}}_2=\frac{{\mathrm{θ}}_1}{2}$

where,${\mathrm{θ}}_1=$angle of incidence

${\mathrm{θ}}_2=$angle of refraction

Substituting the values

$$\begin{gathered} \left(1.00\right)\sin {\mathrm{θ}}_1=n_2\sin \left(\frac{{\mathrm{θ}}_1}{2}\right) \\ \sin 2\left(\frac{{\mathrm{θ}}_1}{2}\right)=n_2\sin \left(\frac{{\mathrm{θ}}_1}{2}\right) \\ \left(u\sin g\sin 2x=2\sin x\cos x\right) \\ 2\sin \left(\frac{{\mathrm{θ}}_1}{2}\right)cos\left(\frac{{\mathrm{θ}}_1}{2}\right)=n_2\sin \left(\frac{{\mathrm{θ}}_1}{2}\right) \\ n=2cos\left(\frac{{\mathrm{θ}}_1}{2}\right) \\ {\mathrm{θ}}_1={\cos }^{-1}\left(\frac{n}{2}\right) \end{gathered}$$

We are given that We are given that the refractive index of air $n_{air}=1.00$and refractive index of glass is $n_{glass}=1.50$. We have to find angle of incidence for glass

From part(a) We can write

$$\begin{gathered} {\mathrm{θ}}_1={\cos }^{-1}\left(\frac{1.50}{2}\right) \\ {\mathrm{θ}}_1=82.8^{°} \end{gathered}$$

Where ${\mathrm{θ}}_1=$angle of incidence for glass