The light ray has an angle of 37.6 o a) Determine the refracted angle in materia
ID: 1562267 • Letter: T
Question
The light ray has an angle of 37.6o
a) Determine the refracted angle in material “a. n=1.36”
b) Determine the angle the light ray makes in material “a n=1.36” when it encounters material “b n=1.56” and find the refracted angle in material “b. n=1.56”
c) Determine the angles for light traveling through the remaining material (c n=1.46), then into air.
d) Create a sketch of Figure 1 on separate paper showing the incoming and refracted rays as it travels through the three layers. Make sure to clearly indicate all angles light makes at each boundary.
Explanation / Answer
LEt consider the ligt ray from air to a material of refractive index n = 1.36
from Snell's law )
na sin theta a = nb sin theta b
thetab1 = arc sin (na sin theta a / nb)
= arc sin (1 sin37.6 /1.36)
= 26.65 degrees
now it is the angle of incidence when light entering from n= 1.36 to the material with n = 1.56
from Snell's law
theta b2= arc sin (na sin theta a / nb)
= arc sin (1.36 sin26.65 /1.56)
= 23.01 degrees
and when the light entering from the material with n = 1.56 to n= 1.46 , the angle of refraction is
from Snell's law
theta b3= arc sin (na sin theta a / nb)
= arc sin (1.56 sin23.01 /1.46)
= 24.69 degrees
finally the light ray from n = 1.46 to n=1 material
the angle of refraction is
from Snell's law
theta b4= arc sin (na sin theta a / nb)
= arc sin (1.46 sin24.69 /1)
= 37.58 degrees
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.