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

157077501.7 m/s

Step-by-step explanation:

Refractive index of water = n₂ = 1.33

θ₁ = 14.5°

θ₂ = 21.8°

According to Snell's law

`(sin\theta_1)/(sin\theta_2)=(n_2)/(n_1)\\\Rightarrow n_1* sin\theta_1=n_2* sin\theta_2\\\Rightarrow n_1* sin14.5=1.33* sin 21.8\\\Rightarrow n_1=(1.33* sin 21.8)/(sin14.5)=1.972`

Refractive index of plastic is 1.973

`n=(c)/(v)\\\Rightarrow v=(c)/(n)\\\Rightarrow v=(3* 10^8)/(1.973)=157077501.7\ m/s`

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