At room temperature, if the relative permittivity of water is 80 and the relative permeability is 0.022, then the velocity of light in water is m s⁻¹.

Question

Accepted Answer

(3) 2.25×10⁸. Given: εᵣ = 80 and μᵣ = 0.022, as the booklet prints them. Idea: in a medium v = 1/(√με) = c/(√μᵣεᵣ), so all that is needed is n = √μᵣεᵣ. μᵣεᵣ = 80×0.022 = 1.76, so n = √1.76 = 1.33. v = (3×10⁸)/(1.33) = 2.26×10⁸ m s⁻¹. To the two figures the options carry, that is 2.25×10⁸ m s⁻¹ -- the speed of light in water, whose refractive index is 1.33.

At room temperature, if the relative permittivity of water is 80 and the relative permeability is 0.022, then the velocity of light in water is ______ m s⁻¹.

Step-by-step solution

Why the other options are wrong