Question

Light of wavelength 500nm traveling with a speed of $2.0 \times 10^8 ms^{-1}$ in a certain medium enters another medium of refractive index 5/4 times that of the first medium. What are the wavelength and speed in the second medium?
A) $400nm, 1.6 \times 10^8 ms^{-1}$
B) $400nm, 2.5 \times 10^8 ms^{-1}$
C) $500nm, 2.5 \times 10^8 ms^{-1}$
D) $625nm, 1.6 \times 10^8 ms^{-1}$

Answer 1

Use the formula of refractive index in the form which contains the relation between the refractive index, wavelength, and speed. On making one equation in terms of wavelength and refractive index and another equation in terms of speed and refractive index of the medium we can easily determine the value of wavelength and speed in the second medium.

Complete step by step solution:
We know that in optics, the refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as where c is the speed of light in vacuum and v is the phase velocity of light in the medium.
General formulas of refractive index are given by
$\mu {\text{ = }}\dfrac{{{v_{\text{1}}}}}{{{v_{\text{2}}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$
$\mu {\text{ = }}\dfrac{{{\lambda _{\text{1}}}}}{{{\lambda _{\text{2}}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)$
Where
$v_1$ and $v_2$ = the velocities of light in two different mediums
${\lambda _{\text{1}}}{\text{ and }}{\lambda _{\text{2}}}$= wavelength of respective mediums
$\mu$= Refractive index of medium
Here, given
${\lambda _{\text{1}}}{\text{ = 500 nm}}$
$\mu {\text{ = }}\dfrac{5}{4}$
Now, putting the values in equation (1) we have
$\dfrac{{\text{5}}}{{\text{4}}}{\text{ = }}\dfrac{{{\text{500}}nm}}{{{\lambda _{\text{2}}}}}$
On solving for ${\lambda _{\text{2}}}$ we get
${\lambda _{\text{2}}}{\text{ = 400}}nm$
Hence the required wavelength of the second medium is 400nm.
Now, for finding the value of speed we use following relation of refractive index with speed: -
Here, given
$v_1= 2.0 \times 10^8 ms^{-1}$
$\mu {\text{ = }}\dfrac{5}{4}$
Now, on putting values in equation (2) we have
$\dfrac{{\text{5}}}{{\text{4}}}{\text{ = }}\dfrac{{{{\;2}}{\text{.0}} \times {\text{1}}{{\text{0}}^{\text{8}}}m{s^{ - 1}}}}{{{v_{\text{2}}}}}$
On solving for ${v_{\text{2}}}$ we have
${v_{\text{2}}}{{ = \;1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{\text{8}}}m{s^{ - 1}}$

$\therefore$ The required speed of light in second medium is ${\text{1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{\text{8}}}m{s^{ - 1}}$. Hence, Option (A) is correct.

Note:
Use the correct relation between refractive index, wavelength, and speed. Also keep in mind that while traveling from one medium to another the frequency of light remains constant because it is the basic property of light but the speed and wavelength of the medium depends on the medium in which the light is traveling.