Question

The critical angle for light going from a medium in which wavelength is $4000{{A}^{o}}$ to medium in which its wavelength is 6000$$$${{\alpha }^{o}} is
A) ${{30}^{\circ }}$
B)${{45}^{\circ }}$
C)${{60}^{\circ }}$
D)${{\sin }^{-1}}\left( 2/3 \right)$

Answer 1

Hint : By using the relation between wavelength, velocity and refractive index we can find the ratio of refractive index. Since the ratio of velocity will be equal to the ratio of wavelength which is equal to the reciprocal of the ratio of refractive index. Then we should apply the relation of refractive index and critical angle.

Complete step-by-step solution:
Let us assume the Refractive index of medium 1 is ${{\mu }_{1}}$
Let us assume the Refractive index of medium 2 is ${{\mu }_{2}}$
When light passes from one medium to another then it will bend or go away from the normal then according to Snell’s law.
$\mu _{2}^{1}=\dfrac{\sin i}{\sin r}$
Where
${{\sin }^{{}}}i$ is defined as the angle of incidence.
${{\sin }^{{}}}r$is defined as the angle of refraction.
Refractive index and velocity are related to each other as velocity and refractive index are inversely proportional to each other and velocity is directly proportional to wavelength of light so wavelength is also inversely proportional to refractive index.

${{v}_{1}}$ is representing the velocity of light in medium 1
${{v}_{2}}$is representing the velocity of light in medium 2.
When light passes from medium 1 to medium 2 then refractive index and velocity are mathematically related as
$_{1}{{\mu }_{2}}=\dfrac{{{v}_{1}}}{{{v}_{2}}}$
And we know that
$velocity=frequency\times wavelength$
Since frequency of light is same in refraction and reflection both. So velocity is directly proportional to wavelength.
So velocity, wavelength and refractive index are mathematically related as:-
$_{1}{{\mu }_{2}}=\dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}$
$\Rightarrow \dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}=\dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}$
Put the values of wavelength from the question we get,
$\dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}=\dfrac{4000{{\alpha }^{o}}}{6000{{\alpha }^{o}}}$
$\therefore \dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}=\dfrac{2}{3}$(Equation 1)
Here, we get the ratio of the refractive index of medium 2 to medium 1.
For critical angle formation light must pass from denser to the rarer medium at that particular angle of incidence in the denser medium at which the angle of refraction becomes ${{90}^{0}}$in the rarer medium.
Since in this question light is passing from the medium 1 where wavelength is $4000{{A}^{o}}$ to medium 2 in which its wavelength is 6000$$$${{\alpha }^{o}}.So medium 1 behave as denser medium and medium 2 behaves as rarer medium.
So for calculating critical angle light is passing from medium 1 to medium 2.
Using, Snell law
$_{1}{{\mu }_{2}}=\dfrac{\sin i}{\sin r}$
${{\Rightarrow }_{1}}{{\mu }_{2}}=\dfrac{\sin {{i}_{c}}}{\sin {{90}^{\circ }}}$
$\Rightarrow \dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}=Sin{{i}_{c}}$
Put the value from equation 1, we get

& Sin{{i}_{c}}=\dfrac{2}{3} \\\ & \therefore {{i}_{c}}=Si{{n}^{-1}}(\dfrac{2}{3}) \\\ \end{aligned}$$ Correct option is D. **Note:** Light has three properties: velocity , wavelength and frequency. Velocity depends on the nature of medium and wavelength is directly proportional to the velocity so wavelength also depends on the nature of medium and frequency is independent of the nature of medium it depends on the source.