Question

A ray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) at an angle of 25.5 ◦ with the normal. part of the light is reflected and part is refracted. find the angle between the reflected and the refracted light. answer in units of ◦ .

Answer 1

Answer: $135.29\º$

The figure attached will be helpful to understand the solution.

Here we see two cases, reflection and refraction of light.

According to the laws of reflection:

1. The incident ray, the reflected ray and the normal are in the same plane.

2. The angle of incidence is equal to the angle of reflection.

*Note the normal is perpendicular to the plane, with a $90\º$ angle with the surface

And this can be visualized in the figure, where the angle ${\theta}_{1}$ with which the incident ray hits the surface is equal to the angle with which this same ray is reflected.

On the other hand we have Refraction, a phenomenon in which  the light bends or changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index is a number that describes how fast light propagates through a medium or material.  

According to Snell’s Law:  

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$     (1)  

Where:  

$n_{1}=1$ is the first medium refractive index  (the air)

$n_{2}=1.309$ is the second medium refractive index  (the ice)

$\theta_{1}=25.5\º$ is the angle of the incident ray  

$\theta_{2}$ is the angle of the refracted ray  

Now, firstly we have to find $\theta_{2}$ and then, by geometry, find $\alpha$ and $\beta$ which sum the angle between the reflected and the refracted light.

Finding $\theta_{2}$ from (1):

$(1)sin(25.5\º)=(1.309)sin(\theta_{2})$    

$sin(\theta_{2})=0.328$

$\theta_{2}=arcsin(0.328)$

$\theta_{2}=19.201\º$   (2)

Remembering that the Normal makes  a $90\º$ angle with the surface, we can say:

$90\º=\theta_{1}+\beta$    (3)

Finding $\beta$:

$\beta=90\º-25.5\º=64.5\º$    (4)

Doing the same with  $\theta_{2}$ and  $\alpha}$:

$90\º=\theta_{2}+\alpha$    (5)

Finding $\alpha$:

$\alpha=90\º-19.201\º=70.79\º$    (6)

Adding both angles (4) and (6):

$\alpha+\beta=70.79\º+64.5\º$    (7)

$\alpha+\beta=135.29\º$>>>This is the angle between the reflected and the refracted light.