Question

Critical angle for glass is ${\theta }_1$ and that for water is ${\theta }_2$. The critical angle $\theta$ at the water glass interface is such that :

• A. $\theta <({\theta }_1+{\theta }_2)/2$
• B. ${\theta }_1<\theta <{\theta }_2$
• C. $\theta >{\theta }_2$
• D. $\theta <{\theta }_1$

Answer 1

The correct option is C $\theta >{\theta }_2$

$\sin {\theta }_1=\frac{1}{n_{glass}}$

$\sin {\theta }_2=\frac{1}{n_{water}}$

$\sin \theta =\frac{n_{water}}{n_{glass}}$

So $\theta <{\theta }_1$

and $\theta >{\theta }_2$

Hence option 'C' is correct.