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} $
$\Rightarrow \mu_{g}=\frac{1}{\sin i}$