Question

If a critical angle of 60 degree and the refractive index of the first medium is 1.732,the refractive index of second medium is

Answer 1

sin i / sin r = μ₂ /  μ₁

sin 60 / sin 90 =  μ₂/1.732

√3 / 1 = μ₂/ √3

Therefore, μ₂ = √3 x √3 = 3

Answer 2

The refractive index of the second medium is 1.5.

Explanation:

Critical angle, $\theta=60^{\circ}$

The refractive index of the first medium, $n_1=1.732$

It is required to find the refractive index of second medium. Using Snell's law as :

$\sin\theta=\dfrac{n_2}{n_1}\\\\n_2=\sin\theta\times n_1\\\\n_2=\sin(60)\times 1.732\\\\n_2=1.499\\\\n_2=1.5$

So, the refractive index of the second medium is 1.5.

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Critical angle

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