Light is incident on a prism with index of refraction, n=1.55, striking the midp

Question

Light is incident on a prism with index of refraction, n=1.55, striking the midpoint of the front surface as shown.

a) What angle (theta1) does the light make with the normal to front surface of the prism once inside the prism?

b) What is the critical angle for the prism for total internal reflection of light when it reaches the glass/air interface when exiting the prism?

c) What surface will the light leave the prism through, front, back, or bottom and at what angle relative to the normal of this surface?

d) How many total internal reflections will the light undergo before leaving the prism?

Explanation / Answer

a)for front n1*sink1=n2*sink2, where n1 =1 (RI of air is 1),is the R.I of outside and n2 =1.55, is RI of inside.

sink1=21 degree , sink2=? therefore sink2=1*sin21/1.55=0.231, where k2 = 13.37 degrees

b)if light is incident on critical angle it refrects by 90 degree. say iincident angle is k3 then sink3*n2=siin90*n1, sink3=1/1.55=0.645 and k3 = 40.18 degree, which is the critical angle for prism

c) since the dimension of prisn is not given then there are two possiblities

first, prism is small enough so that light strike on the back. In this case angle of incident ray (41.4 degree) is greater than critical angle so it will reflect. after reflection it will strike on the bottom with the incident angle of 29.6 degrees and will go out.

second case if the dimensions of prism are large enough light will strike to the bottom with the iincident angle (67.6 degree) greater than the critical angle so reflection will take place. Then it will strike on the Back and will go out side because in this case also incident angle is less than critical angle.

d) In either case light is going in total reflection for one time only

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