Question

A ray of light is incident at the glass-water interface at an angle $i$, it emerges finally parallel to the surface of water, then the value of ${\mu }_g$ would be

• A. $(4/3)\sin i$
• B. $1/\sin i$
• C. $4/3$
• D. $1$

Answer 1

Answer: (B)

For glass-water interface, applying Snell's law
$\frac{\sin i}{\sin r}=\frac{{\mu }_w}{{\mu }_g}$

$\Rightarrow {\mu }_g=\left(\frac{{\mu }_w\sin r}{\sin i}\right)...(i)$
For water air interface,
Angle of incidence in water $=r$
Again, $\frac{\sin r}{\sin 90^{\circ }}=\frac{1}{{\mu }_w}$
$\Rightarrow \sin r=\frac{1}{{\mu }_w}...(ii)$
From Eqs. (i) and (ii),
${\mu }_g=\frac{\left({\mu }_w\right)\times \left(\frac{1}{{\mu }_w}\right)}{\sin i}$
$\Rightarrow {\mu }_g=\frac{1}{\sin i}$