Download Solutions #2

Physics 370 - Optics
Homework Assignment #2
Chapter 2: #5, 7
Problem 2 - 5 :
ü A ray of light makes an angle of incidence of 45° at the center of the top surface of a transparent
cube of index 1.414. Trace the ray through the cube.
Solution : I am assuming that the ray travels in a plane
parallel to two of the cube ' s faces. This is the simplest case.
Part 1 – As the ray is refracted at the first surface
Hassuming the cube is surrounded by airL,
the angle of the ray is determined by Snell ' s Law : n1 sin q1 =
n2 sin q2 . We find that q2 = 30 °.
n1 = 1.0; n2 = 1.414; q1 = 45 Degree;
1
q2 = ArcSinB
n1 Sin@q1DF;
n2
Print@"Angle of refraction = ", q2 ê Degree, "°"D
Angle of refraction = 30.005°
Part 2 – The ray travels through the cube approximately toward one the
bottom edges. We must determine if it strikes a vertical face or the
bottom face. We will calculate the lateral distance travelled when
the vertical distance is equal to the length of one edge of the cube.
The lateral displacement will be larger than half the length of one side =
distance from the center of one face to the edge.
displacement = side * Tan@q2D
0.577467 side
Therefore the ray will hit one side,
be total internally reflected and exit from the bottom of the cube at 45 °,
a short distance from the edge. This distance can be calculated by determining
exactly where the ray strikes the side of the cube. @This part not completedD
Problem 2 - 7 :
ü A small source of light at the bottom face of a rectangular glass slab 2.25 cm thick is viewed
from above. Rays of light totally reflected at the top surface outline a circle of 7.60 cm diameter
on the bottom surface. Determine the refractive index of the glass.
2
Homework_Ch_02a.nb
A small source of light at the bottom face of a rectangular glass slab 2.25 cm thick is viewed
from above. Rays of light totally reflected at the top surface outline a circle of 7.60 cm diameter
on the bottom surface. Determine the refractive index of the glass.
Solution :
The critical angle can be calculated from the given dimensions. H7.60 ê 4 L cm is the
length of the bottom side of the triangle and 2.25 cm is the height of the triangle,
so the critical angle can be calculated. If the medium outside the
1
glass has an index of refraction of 1.0, then
= sin qc .
n
qc = ArcTanB
2.25
F ; H* degrees *L
7.60 ê 4
Print@"Critical angle = ", qc ê Degree, "°"D
Critical angle = 49.8208°
n=
1
;
Sin@qcD
Print@"Index of refraction = ", nD
Index of refraction = 1.30885
ü