Question

In the figure ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of $\theta$ so that light incident normally on the face AB does not cross the face BC is (given sin${}^{-1}$ (3/5) $=$ 37${}^o$)

• A. $\theta \le {37}^o$
• B. $\theta >{37}^o$
• C. $\theta \le {53}^o$
• D. $\theta <{53}^o$

Answer 1

Answer: (A)

$\theta \le {37}^o$

For no refraction at surface BC

$\frac{sin\left(i\right)}{sin\left(r\right)}=\frac{{\mu }_2}{{\mu }_1}fornorefractionatsurfaceBC$

$r\ge {90}^0$

here,

${\mu }_1=\frac{3}{2}\left(refractiveindexofABC\right)$

${\mu }_2=\frac{6}{5}\left(refractiveindexofBCDE\right)$

thus

$\frac{sin\left(i\right)}{sin\left(90\right)}=\frac{\frac{6}{5}}{\frac{3}{2}}=\frac{4}{5}$

$\Rightarrow sini\ge \frac{4}{5}$

$\Rightarrow i\ge {53}^0$

here in triangle ABC

$i={90}^0-\theta \Rightarrow {90}^0-\theta \ge {53}^0\theta \le 37^0$