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Question: What is the speed of light in a denser medium of Polarising angle \(30\)?...

What is the speed of light in a denser medium of Polarising angle 3030?

Explanation

Solution

It says that the tangent of the polarizing angle is numerically equivalent to the refractive index of the system. A polarizing angle makes when at a particular incidence angle, the reflected light is polarized, and this particular value of the incidence angle is identified as the polarizing angle. The refractive index is equal to the light speed divided by the light speed through the medium.

Complete step by step answer:
Given: Polarising angle, θ=30\theta = 30^{\circ}
The polarising angle and refractive index are related by the formula:
μ=tanθ\mu = \tan \theta
Put θ=30\theta = 30^{\circ} in the above formula.
μ=tan30\mu = \tan 30^{\circ}
μ=13\mu = \dfrac{1}{\sqrt{3}}
μ\mu is the refractive index.
θ\theta is the polarising angle.
We have to find the speed of light in a denser medium.
Refractive index is the speed of light in vacuum divided by the speed of light in the denser medium.
μ=cv\mu = \dfrac{c}{v}
    v=cμ\implies v= \dfrac{c}{\mu }
    v=3×3×108\implies v = \sqrt{3} \times 3 \times 10^{8}
    v=5.2×108ms1\implies v = 5.2 \times 10^{8} m s^{-1}
The speed of light in a denser medium is 5.2×108ms15.2 \times 10^{8} m s^{-1}.

Note: As the wave moves into the less solid medium, it promotes up, bending apart from the normal line. The refractive index represents the ratio of the speed in space in connection to the speed of the medium; thus, the speed will be higher in a medium with a feebler refractive index.