Question

Difficulty: EasyWave Phenomena: Reflection, Refraction, Interference, Diffraction, and Polarization

A ray of light travels from air into a liquid with a refractive index of 1.331.33. If the sine of the angle of incidence in air is 0.800.80, what is the sine of the angle of refraction in the liquid?

Answer: 0.6

Answer

The sine of the angle of refraction in the liquid is 0.60.
According to Snell's law for light passing from air into a medium, the refractive index nn is given by n=sinisinrn = \frac{\sin i}{\sin r}. Rearranging this equation to solve for the sine of the angle of refraction yields sinr=sinin\sin r = \frac{\sin i}{n}. Substituting sini=0.80\sin i = 0.80 and n=1.33n = 1.33 (or 43\frac{4}{3}) gives sinr=0.804/3=0.60\sin r = \frac{0.80}{4/3} = 0.60.

Step-by-Step Solution

1
Identify the given physical parameters and state Snell's law
Refractive index n=1.33n = 1.33 (or 43\frac{4}{3}), sini=0.80\sin i = 0.80. Snell's law: n=sinisinrn = \frac{\sin i}{\sin r}
Snell's law relates the ratio of the sines of the angles of incidence and refraction to the refractive index of the medium.
2
Rearrange the equation to express the sine of the angle of refraction
sinr=sinin\sin r = \frac{\sin i}{n}
Algebraically isolating sinr\sin r allows direct substitution of the known quantities.
3
Substitute the values and compute the result
\sin r = \frac{0.80}{4/3} = 0.60
Dividing 0.800.80 by 43\frac{4}{3} gives 0.600.60.

Key Concept

Snell's Law of Refraction

Alternative Method

Convert decimal numbers into simple fractions: n=43n = \frac{4}{3} and sini=45\sin i = \frac{4}{5}. Evaluating sinr=4/54/3\sin r = \frac{4/5}{4/3} simplifies directly to 35=0.60\frac{3}{5} = 0.60.
Estimated Time:45s
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