The refractive indices of glycerine and diamond with respect to air are 1.4 and 2.4 respectively. Calculate the speed of light in glycerine and diamond. From these results, calculate the refractive index of diamond with respect to glycerine. (c $= 3 \times 10^{8}$ m/s)

2.143 \times 10^{8}

m/s;

1.250 \times 10^{8}

m/s; 1.714

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0.2143 \times 10^{8}

m/s;

12.50 \times 10^{5}

m/s; 0.1714

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2.143 \times 10^{6}

m/s;

1.250 \times 10^{3}

m/s; 1.74

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21.43 \times 10^{8}

m/s;

0.125 \times 10^{5}

m/s; 17.14

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Solution

The correct option is A $2.143 \times 10^{8}$ m/s; $1.250 \times 10^{8}$ m/s; 1.714
Refractive index, $1.4 = \frac{c}{v_{1}} = \frac{3 \times 10^{8}}{v_{1}}$

$v_{1} = 2.14 \times 10^{8} m / s$

Also, $2.4 = \frac{3 \times 10^{8}}{v_{2}}$

$v_{2} = 1.25 \times 10^{8} m / s$

Refractive index of diamond wrt to glycerine, $μ = \frac{2.4}{1.4} = 1.714$

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