Question

In your laboratory, you trace the path of light rays through a glass slab for different values of angle of incidence $\left(\angle i\right)$ and in each case, measure the values of the corresponding angle of refraction $\left(\angle r\right)$ and angle of emergence$\left(\angle e\right)$. Based on your observations, the correct conclusion is

• A. $\angle i$ is more than $\angle r$ , but nearly equal to $\angle e$
• B. $\angle i$ is less than $\angle r$ , but nearly equal to $\angle e$
• C. $\angle i$ is more than $\angle e$ , but nearly equal to $\angle r$
• D. $\angle i$ is less than $\angle e$ , but nearly equal to $\angle r$

Answer 1

The correct option is A $\angle i$ is more than $\angle r$ , but nearly equal to $\angle e$

Since the light ray enters from a rarer (air) medium to a denser (glass) medium, the ray bends toward normal. As a result, the angle of incidence (angle between the incident ray and normal) $\angle i$ is more than the angle of refraction (angle between refracted ray and normal ) $\angle r$.

Let $n_2$ be the refractive index of denser medium(glass slab).

Let $n_1$ be the refractive index of rarer medium(air).

Now using Snell's law, when the light ray goes into glass from the air,

$\frac{n_2}{n_1}=\frac{\sin i}{\sin r}$ ...................$\left(1\right)$,

When the light ray goes into the air from glass,

$\frac{n_1}{n_2}=\frac{\sin r}{\sin e}$

or $\frac{n_2}{n_1}=\frac{\sin e}{\sin r}$ ........................$\left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$ , we get ,

$\frac{\sin i}{\sin r}=\frac{\sin e}{\sin r}$ ,

$\sin i=\sin e$

or $\angle i=\angle e$

Thus, $\angle i$ is more than $\angle r$, but nearly equal to $\angle e$.