Question

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

• A. $\left(\frac{4}{3}\right)sin\left(i\right)$
• B. $\frac{1}{sin \left(i\right)}$
• C. $\frac{4}{3}$
• D. $1$

Answer 1

Answer: (B)

Applying, Snell's law $\left(μsin \left(i\right) = \text{ constant }\right)$ at $1$ and $2$ , we have

$ \mu_{1} \sin i _{1}=\mu_{2} \sin i _{2} $
Here, $\mu_{1}=\mu_{\text {glass }}, \quad i_{1}= i$
$\therefore \mu_{2}=\mu_{\text {air }}=1$ and $i _{2}=90^{\circ}$
$ \therefore \mu_{ g } \sin ( i )=1 \times \sin 90^{\circ} $
$ \therefore \quad \mu_{ g }=\frac{1}{\sin ( i )} $