Question

Monochromatic light enters an equilateral
prism made of glass with index of refraction
1.42 in such a way that, after refraction at the
first surface, the light travels parallel to the
base of the prism,
30
o
What must be the angle of incidence of the
light for this to occur?
Answer in units of
◦

Answer 1

Monochromatic light enters an equilateral prism made of glass with index of refraction 1.42 in such a way that, after refraction at the first surface, the light travels parallel to the base of the prism,

30

o

What must be the angle of incidence of the light for this to occur?

Answer in units of

o

Solution:

According to Snell's law:

$$n = \frac {\sin \theta_ {1}}{\sin \theta_ {2}}$$

were $\theta_{1}$ - the angle of incidence, $\theta_{2}$ - the angle of refraction, $n$ - refraction index

$$\theta_ {2} = 3 0 {}^ {\circ}$$ $$n = 1. 4 2$$ $$\sin \theta_ {1} = n \sin \theta_ {2}$$ $$\theta_ {1} = \arcsin (n \sin \theta_ {2}) = \arcsin (1. 4 2 * \sin 3 0 {}^ {\circ}) = 4 5. 2 3 {}^ {\circ}$$

Answer: $45.23{}^{\circ}$ or $45.23 - 30 = 15.23{}^{\circ}$ to the base of the prism.