Physics, 14.04.2020 19:47 mal5546

A ray of light strikes the midpoint of one face of an equiangular (60°–60°–60°) glass prism (n = 1.5) at an angle of incidence of 39.8°. (a) Trace the path of the light ray through the glass, and find the angles of incidence and refraction at each surface.

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Answer from: invocx

The path of light ray through the glass is shown on the first uploaded image

First surface:

Angle of incidence is i_1 = 39.0^o

Angle of refraction is r_1= 25.23^o

Second surface:

Angle of incidence is i_2 = 34.77^o

Angle of refraction is r_2= 58.80^o

Since the angle of incidence is equal to the angle of reflection

Then at the first surface the angle of reflection is R = 39.8^o

And at the first surface the angle of reflection is R_2 = 34.77^o

Explanation:

From the question we are told that

The angle of incidence is i_1 = 39.0^o

The refractive index of the prism is $n_{p} = 1.5$

The angle of the prism is A = 60^o

The path of light ray through the glass is shown on the first uploaded image

For the first surface of the prism

According to Snell's law

$n_{air} sin( i_1) = n_{p} sin( r_1)$

The refractive index of air $n_{air}$ has a constant value of 1

Now making the angle of refraction at the first surface of the prism r_1 the subject

$r_1 = sin^{-1}[\frac{sin (i_1) }{n_{p}} ]$

$= sin^{-1}[\frac{sin(39.8)}{1.5} ]$

r_1= 25.23^o

For the second surface of the prism

looking at the diagram on the first uploaded image the angle of incidence is mathematically evaluated as

i_2 = A - r_1

Substituting values

i_2 = 60 -25.23

=34.77^o

According to Snell's law

$n_{p} sin( i_2) = n_{air} sin( r_2)$

Now making the angle of refraction at the second surface of the prism r_2 the subject

$r_2 = sin^{-1} [\frac{n_p sin(i_2)}{n_air} ]$

Substituting values into the equation

$r_2 = sin^{-1} [\frac{1.5 * sin(34.77)}{1}]$

r_2 =58.8^o

Answer from: Quest

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Answer from: Quest

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A ray of light strikes the midpoint of one face of an equiangular (60°–60°–60°) glass prism (n = 1.5...

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