Need help with angle of incidence of light through a prism

[batcave1985]

Homework Statement

The direction of the ray of light enters equilateral 60° glass prism normal to the surface as shown in attached image
refracted index of the prism material for the violet light is 1.651

(a) What is the angle of incidence on the surface A-B?
(b) Does total internal reflection occur?
(c) Draw a ray diagram to show how light passes through
the prism

Homework Equations

critical angle-----sin i c= 1/n

The Attempt at a Solution

My attempt
a) critical angle
sin= 1/1.651
= 37.3°
Not sure if its 90° for the incidence for side A-B

b) Total internal reflections occur because the angle of incidence is greater
then the critical angle

c) i also attached my attempt to the image.


thank you any help would be great

 

[dauto]

The calculation seems correct, but the way you expressed it is confusing.
You wrote
"sin= 1/1.651
= 37.3°"

You should have written
"sin θ_c= 1/1.651
θ_c = 37.3°"

 

[batcave1985]

The calculation seems correct, but the way you expressed it is confusing.
You wrote
"sin= 1/1.651
= 37.3°"

You should have written
"sin θ_c= 1/1.651
θ_c = 37.3°"

thank you any chance you can tell me if my ray diagram in the picture attached is correct?

 

[ehild]

thank you any chance you can tell me if my ray diagram in the picture attached is correct?

No.

Note that the angle of incidence is defined as the angle the ray encloses with the normal of the surface.
At totalreflecton, the ray is reflected from the surface AB inside the prism, and is incident onto an other side, and then leaves the prism.

ehild

 

[HalfLostAndSearching]

a) The angle of incidence on the surface A-B can be calculated using the law of refraction, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two materials. In this case, we have:

sin i / sin r = n2 / n1

Where n1 is the refractive index of air (approximately 1) and n2 is the refractive index of the prism material (1.651 for violet light). Therefore:

sin i / sin 60° = 1.651 / 1

sin i = sin 60° * (1.651 / 1) = 1.651 * 0.866 = 1.431

i = arcsin(1.431) = 55.4°

So the angle of incidence on the surface A-B is approximately 55.4°.

b) Total internal reflection will occur when the angle of incidence is greater than the critical angle, which can be calculated using the equation you provided:

sin i_c = 1/n

For violet light, the critical angle is therefore:

sin i_c = 1/1.651 = 0.606

i_c = arcsin(0.606) = 37.3°

Since the angle of incidence on surface A-B is greater than the critical angle (55.4° > 37.3°), total internal reflection will occur.

c) I have attached a ray diagram below to show how light passes through the prism. The incident ray enters the prism at an angle of 55.4° and is refracted towards the normal. It then undergoes total internal reflection at the surface B-C, with the angle of reflection being equal to the angle of incidence. Finally, the refracted ray exits the prism at an angle of 55.4° and is bent away from the normal.

A
/ \
/ \
i r
\ /
\ /
B
/ \
/ \
r i
\ /
\ /
C