Question

A plastic cube is immersed in water having an index of refraction of 1.33. A laser beam in the plastic strikes the interface at an angle of 14.5° with the normal in the plastic. In the water, this beam makes an angle of 21.8° with the normal. What is the speed of light in the plastic cube? (c = 3.0 × 108 m/s)

Answer 1

Answer:

157077501.7 m/s

Explanation:

Refractive index of water = n₂ = 1.33

θ₁ = 14.5°

θ₂ = 21.8°

According to Snell's law

$\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}\\\Rightarrow n_1\times sin\theta_1=n_2\times sin\theta_2\\\Rightarrow n_1\times sin14.5=1.33\times sin 21.8\\\Rightarrow n_1=\frac{1.33\times sin 21.8}{sin14.5}=1.972$

Refractive index of plastic is 1.973

$n=\frac{c}{v}\\\Rightarrow v=\frac{c}{n}\\\Rightarrow v=\frac{3\times 10^8}{1.973}=157077501.7\ m/s$

∴ Speed of light in the plastic cube is 157077501.7 m/s