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At room temperature, if the relative permittivity of water is 80 and the relative permeability is 0.0222, then the velocity of light in water is ______ m s⁻¹.

Asked in GSEB Board March 2018 · Speed in vacuum and in a medium

Answer: (3) 2.25×10⁸

Step-by-step solution

Given: εᵣ = 80 and μᵣ = 0.0222 for water at room temperature.

Idea: in a medium v = 1/(√με) = c/(√μᵣεᵣ), so all that is needed is the refractive index n = √μᵣεᵣ.

μᵣεᵣ = 80×0.0222 = 1.776, so n = √1.776 = 1.33.

v = (3×10⁸)/(1.33) = 2.25×10⁸ m s⁻¹.

This is the familiar result for water, whose refractive index is 1.33: light in it travels at about three quarters of its vacuum speed.

Why the other options are wrong

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