Question

A light beam passing from medium 2 to medium 1 is refracted as shown in FIGURE Q34.4. Is n1 larger than n2, is n1smaller than n2, or is there not enough information to tell? Explain

Short answer

The light is refracted as n1 is smaller than n2

Step-by-step solution

Step1:snell"s law

Snell's Law: It relates the refraction angle ${\theta }_2$to the incident angle ${\theta }_1$and the indexes of refraction of the incident medium n1and the refracted medium n2:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Angles of incidence and refraction are measured with respect to the normal line at the point where the ray intersects the interface between the two media. Medium 2 is more optically dense (has a higher refractive index) than medium 1 if the refracted ray is closer to the normal than the incident ray. In the same plane are the incident ray, refracted ray, and normal line.

Given Data

The index of refraction of medium 1 is: n1

The index of refraction of medium 2 is:n2

Step3:Find n1 larger than n2

We are asked to determine if n1 is larger than n2, n1 smaller than n2, or if there is not enough information to tell.

Step4:Apllying snell's law

Let ${\theta }_1$be the angle of the light ray with the normal to the surface of the medium and ${\theta }_2$be the angle of the light ray with the normal the surface of the medium. Using this notation, we clearly see in the given figure that ${\theta }_1>{\theta }_2$which also means that $\sin {\theta }_1>\sin {\theta }_1$.

$\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{n_2}{n_1}$

Hence, if $\sin {\theta }_1>\sin {\theta }_2,n_1$must be smaller than n2