Question

The refractive index of denser medium with respect to rarer medium is 1.125. difference between the velocities of light in the two media is $0.25{10}^8$ m/s find the velocities of light in the two media and their refractive indices $(c=3\times {10}^8)$

• A. $2.0\times {10}^{8}m/s;2.25\times {10}^{8}m/s;1.500;1.333$
• B. $2.5{10}^{8}m/s;2.25\times {10}^{8}m/s;1.500;1.8888$
• C. $2\times {10}^{8}m/s;2.25\times {10}^{8}m/s;1.333;1.880$
• D. $2.25\times {10}^{8}m/s;2\times {10}^{8}m/s;1.500;1.333$

Answer 1

The correct option is A $2.0\times {10}^{8}m/s;2.25\times {10}^{8}m/s;1.500;1.333$

Given,

The refractive index of a denser medium ${\mu }_d$ with respect to a rarer medium ${\mu }_r$ is given as

$\frac{{\mu }_d}{{\mu }_r}=1.125$

And the difference between their velocities given as

$v_r-v_d=0.25\times {10}^{8}m/s$ …… (1)

The ratio of the refractive index is given as

$\frac{{\mu }_d}{{\mu }_r}=\frac{v_r}{v_d}=1.125$

$v_r=1.125v_d$ …… (2)

Substitute the value of equation (2) in (1)

$1.125v_d-v_d=0.25\times {10}^{8}m/s$

$0.125v_d=0.25\times {10}^{8}m/s$

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

Hence from equation (2) the value of $v_d$ is

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

Now, as the refractive index of the denser medium can be given as

${\mu }_d=\frac{c}{v_d}$

Where $c=3\times {10}^{8}m/s$ is the velocity of light in a vacuum

${\mu }_d=\frac{3\times {10}^{8}m/s}{2\times {10}^{8}m/s}$

${\mu }_d=1.5$

Hence the refractive index of the rarer medium can be obtained as

$\frac{{\mu }_d}{{\mu }_r}=1.125$

${\mu }_r=1.333$