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

Answer 1

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

Given - critical angle for glass $={\theta }_1$

critical angle for water $={\theta }_2$

critical angle for water glass interface $=\theta$

for glass, the critical angle is

$\sin {\theta }_1=\frac{1}{{}_a{\mu }_g}$

for water, the critical angle is

$\sin {\theta }_2=\frac{1}{{}_a{\mu }_w}$

as we know ${}_a{\mu }_g{>}_a{\mu }_w$

hence $\sin {\theta }_2>\sin {\theta }_1$

for water glass interface ,the critical angle is

$\sin \theta =\frac{1}{{}_w{\mu }_g}$

or $\sin \theta =\frac{{}_a{\mu }_w}{{}_a{\mu }_g}=\frac{\sin {\theta }_1}{\sin {\theta }_2}$

but $\sin {\theta }_2>\sin {\theta }_1$

therefore $\sin \theta >\sin {\theta }_2$

or $\theta >{\theta }_2$