Question

4. A diamond is immersed in a liquid with a refractive index greater than
water. Then the critical angle for total internal reflection will
(a) will depend on the nature of the liquid
(b) decrease
(c) remains the same
(d) increase​

Answer 1

(b) it will decrease.

Answer 2

D) The critical angle will increase

Critical Angle

• When the light travels from a denser medium to a rarer medium, the angle of incidence at which the angle of refraction becomes $90^{o}$ is said to be the critical angle.  

• When the angle of incidence increases beyond the critical angle, the light gets reflected back to the denser medium. This phenomenon is called Total Internal Reflection (TIR).

Calculating critical angle

Snell's Law of refraction is,

$n_{1} sin(\theta_{1}) \ = \ n_{2} sin(\theta_{2})$

Where, $n_{1}$ & $n_{2}$ are refractive indices of the rarer and denser medium, respectively. $\theta_{1}$ and $\theta_{2}$ are the angle of incidence and angle of refraction, respectively.

Consider $\theta_{1}$ as the critical angle while $\theta_{2}$ is the angle of refraction, which is $90^{o}$.

Therefore,

$\theta_{c} \ = \ sin^{-1} (\frac{n_{2}}{n_{1}})$

With the above expression, we can say that, if the refractive index of the liquid is increased, then the angle $\theta_{c}$ should be increased as $\theta_{c}$ is proportional to $n_{2}$.

For example,

• If ${n_{2}$ = 1.33 (refractive index of water) and $n_{1}$ = 2.42 (refractive index of diamond), $\theta_{c}$ will be $33.339^{o}$.

• And, if we increase ${n_{2}$ to 1.39 (refractive index of kerosene), then $\theta_{c}$ will become $35.056^{o}$.

Therefore, the critical angle will increase with the increase in the refractive index of the rarer medium. But, the refractive index of the rarer medium should not exceed the value of the denser one as it will violate the concept of critical angle and Total Internal Reflection.