Question

The critical angle for glass is ${41}^{\circ }{48}^{\prime }$ and that for water is ${48}^{\circ }{36}^{\prime }$. Calculate the critical angle for glass-water interface.

• A. ${62}^{\circ }{28}^{\prime }$
• B. ${34}^{\circ }{42}^{\prime }$
• C. ${52}^{\circ }{42}^{\prime }$
• D. ${44}^{\circ }{42}^{\prime }$

Answer 1

The correct option is A ${62}^{\circ }{28}^{\prime }$

Critical angle $\sin {\theta }_c=\frac{1}{n}$

We get $n=\frac{1}{\sin {\theta }_c}$

We will use the relation $1^o={60}^{\prime }$

For glass :

Critical angle ${\theta }_c={41}^o{48}^{\prime }={41}^o+\frac{48}{60}={41}^o+{0.8}^o={41.8}^o$

Refractive index for glass $n_g=\frac{1}{\sin \left(41.8\right)}=\frac{1}{0.66}=1.5$

For water :

Critical angle ${\theta }_c={48}^o{36}^{\prime }={48}^o+\frac{36}{60}={48}^o+{0.6}^o={48.6}^o$

Refractive index for water $n_w=\frac{1}{\sin \left(48.6\right)}=\frac{1}{0.75}=1.33$

For water-glass interface :

Critical angle $\sin {\theta }_c=\frac{n_w}{n_g}=\frac{1.33}{1.5}=0.887$

$⟹{\theta }_c={\sin }^{-1}\left(0.887\right)={62.46}^o$

We get ${\theta }_c={62}^o+(0.46\times 60{)}^{\prime }={62}^o{28}^{\prime }$