Chapter 18, Problem 80A

To determine

To compute: - The large index of refraction between the medium A and B and explain it.

Answer to Problem 80A

  $n_A>n_B$

Explanation of Solution

Formula used:

According to the Snell’s law:

  $n_A\sin {\theta }_i=n_B\sin {\theta }_r$

Where,

  $n_A=$ Index of refraction of medium A.

  $n_B=$ Index of refraction of medium B.

  ${\theta }_i=$ Angle of incident.

  ${\theta }_r=$ Angle of refraction.

Calculation:

From the given diagram, the angle of incidence is less than the angle of refraction becausethe refracted ray is moving away from the normal.

Since,

  ${\theta }_i<{\theta }_r$

Therefore,

  $\sin {\theta }_i<\sin {\theta }_r$

From Snell’s law:

  $\begin{array}{c}\frac{n_A}{n_B}=\frac{\sin {\theta }_r}{\sin {\theta }_i}\\ \therefore \\ \frac{n_A}{n_B}>1\\ n_A>n_B\end{array}$

The light moving from denser medium to rarer medium. And during the refraction of light, the refracted ray would bend away from the normal.

Conclusion:

Refractive index of medium A is greater than the refractive index of medium B.