The refractive index of water is 1.33. What will be the speed of light in water?

Question

The refractive index of water is 1.33. What will be the speed of light in water? (A)  4 × 108 m/s  (B)  1.33 × 108 m/s  (C)  3 × 108 m/s  (D)  2.25 × 108 m/s

Tardigrade Admin · Accepted Answer

Refractive index of water (μ 2) = 1.33. (v2/v1) = (μ1/μ2) = (1/1.33) therefore v2 = (v1/1.33) = (3 × 108/ 1.33) = 2.25 × 108 m/s

Q. The refractive index of water is 1.33. What will be the speed of light in water?

$4 \times 10^{8} m / s$

$1.33 \times 10^{8} m / s$

$3 \times 10^{8} m / s$

$2.25 \times 10^{8} m / s$

Refractive index of water $(μ_{2})$ = 1.33.
$\frac{v _{2}}{v _{1}} = \frac{μ _{1}}{μ _{2}} = \frac{1}{1.33}$
therefore $v_{2} = \frac{v _{1}}{1.33} = \frac{3 \times 10 ^{8}}{1.33} = 2.25 \times 10^{8} m / s$

Q. The refractive index of water is 1.33. What will be the speed of light in water?

4×108m/s

1.33×108m/s

3×108m/s

2.25×108m/s

Refractive index of water (μ2​) = 1.33. v1​v2​​=μ2​μ1​​=1.331​ therefore v2​=1.33v1​​=1.333×108​=2.25×108m/s

$4 \times 10^{8} m / s$

$1.33 \times 10^{8} m / s$

$3 \times 10^{8} m / s$

$2.25 \times 10^{8} m / s$

Refractive index of water $(μ_{2})$ = 1.33.
$\frac{v _{2}}{v _{1}} = \frac{μ _{1}}{μ _{2}} = \frac{1}{1.33}$
therefore $v_{2} = \frac{v _{1}}{1.33} = \frac{3 \times 10 ^{8}}{1.33} = 2.25 \times 10^{8} m / s$