Question

If the velocity of light in water is $2.25\times {10}^{8}m/s$ ,then its velocity in carbon disulphide will be (refractive index for carbon-disulphide = 1.63 and refractive index for water is 1.333 ):

• A. $1.84\times {10}^{8}m/s$
• B. $2.25\times {10}^{8}m/s$
• C. $2\times {10}^{8}m/s$
• D. $3\times {10}^{8}m/s$

Answer 1

The correct option is A $1.84\times {10}^{8}m/s$

Given,
Velocity of light in water is $2.25\times {10}^{8}m/s$
Refractive index of Carbon disulphide is $1.63$.
We know that refractive index of water is $1.3330$.
Therefore, velocity of light in Carbon disulphide is :
$v_c=\frac{v_w\times {\mu }_w}{{\mu }_c}$
$=\frac{2.25\times {10}^8\times 1.333}{1.63}$
$=1.84\times {10}^{8}m/s$