No, this question isn't abt language arts, mathematics. Im asking u this How are u doing? c:

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 t

sergij07 [2.7K]

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.

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