91Ó°ÊÓ

At the point of refraction, the angle formed by a refracted beam and a line drawn normal to the interface between two media.

Let us recall the given values.

At sunrise and sunset, sunlight enters the Earth's atmosphere at an incidence angle ${90}^{°}$. Consider how the sun passes through three layers of the Earth's atmosphere. The first layer's refractive index is 1.000170, the refractive index of the second layer is 1.000270, and the third layer's refractive index is 1.000293.

Formula used:

From Snell's law of refraction,

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

Here,

• ${\mathrm{θ}}_1$is the angle between the ray and perpendicular in medium 1.
• ${\mathrm{θ}}_2$is the angle between the ray and perpendicular in medium 2.
• $n_1$is the refractive index of the medium 1.
• $n_2$is the refractive index of medium 2.

Let us solve the given problem.

For1st layer:

Rearrange the above expression.

${\mathrm{θ}}_2={\sin }^{-1}\left(\frac{n_1\sin {\mathrm{θ}}_1}{n_2}\right)-------\left(1\right)$

Substitute ${90}^{°}$for ${\mathrm{θ}}_1,1.000170$ for n₁ and 1.000270 for n₂ in equation (1).

role="math" localid="1653396714174" $$\begin{gathered} {\mathrm{θ}}_2={\sin }^{-1}\left(\frac{\left(1.000170\right)\sin \left({90}^{°}\right)}{1.000270}\right) \\ =89.{18}^{°} \end{gathered}$$

For2nd layer:

The angle of refraction for the second layer is,

${\mathrm{θ}}_3={\sin }^{-1}\left(\frac{n_2\sin {\mathrm{θ}}_2}{n_3}\right)$

Substitute for for and for in the above equation.

$$\begin{gathered} {\mathrm{θ}}_3={\sin }^{-1}\left(\frac{\left(1.000170\right)\sin \left(89.{18}^{°}\right)}{1.000293}\right) \\ =89.{09}^{°} \end{gathered}$$

Therefore, the angle of refraction for various layers is $89.{18}^{°}$and$89.{09}^{°}$.