Question

For very small angle of incidence $(i\approx 3^{\circ })$ of a ray of light striking a glass slab as shown in figure.


The correct plot for lateral displacement (d) of ray vs angle of incidence (i) will be:

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The formula for lateral displacement is,
$d=\frac{t\sin (i-r)}{\cos r}$ ... (i)

From Snell's law:

$\frac{\sin i}{\sin r}=\mu$ ...(ii)

for very small angle $i,r$ will be also very small.
$\Rightarrow \sin i\approx i,\sin (i-r)\approx i-r,\cos r\approx 1$

Thus from Eq. (i),

$d=\frac{t(i-r)}{1}=t(i-r)$

$\Rightarrow d=ti[1-\frac{r}{i}]$

from Eq. (ii) $\frac{r}{i}=\frac{1}{\mu }$

$\Rightarrow d=ti[1-\frac{1}{\mu }]$

or, for the given slab and angle of incidence,
it resembles $y=mx$ equation.

$\Rightarrow {}^{\prime }{d}^{\prime }vs{}^{\prime }{i}^{\prime }$ graph will be a straight line passing from origin.

Hence, option (b) is the correct answer.

Why this question? Tip: Use the approximation $\sin \theta \approx \theta$ and $\cos \theta \approx 1$ for very small value of $\theta$