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77julia77 [94]
3 years ago
10

Type of electromagnetic radiation is used to kills cancer cells

Physics
1 answer:
notka56 [123]3 years ago
3 0
B. gamma rays, is right.
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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.

<u>*Note the normal is perpendicular to the plane, with a 90\º angle with the surface</u>

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 <u>angle between the reflected and the refracted light</u>.

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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