Question 77: Ray Optics

Question

A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $$	heta $$iC and the Brewster's angle of incidence is $$	heta $$iB, such that sin$$	heta $$iC/sin$$	heta $$iB = $$\eta $$ = 1.28. The relative refractive index of the two media is :

Accepted Answer

We are tasked with finding the relative refractive index given the relation between the
critical angle
and the
BrewsterÃ¢ÂÂs angle
.
Step 1: Recall the definitions
Critical angle
(from denser medium $n_1$ to rarer medium $n_2$, with $n_1 > n_2$):
$ \sin 	heta_c = \frac{n_2}{n_1} = \frac{1}{\mu} $
where $\mu = \frac{n_1}{n_2}$ is the relative refractive index $> 1$.
BrewsterÃ¢ÂÂs angle
:
$ 	an 	heta_B = \frac{n_2}{n_1} = \frac{1}{\mu} $
Step 2: Given condition
$ \frac{\sin 	heta_c}{\sin 	heta_B} = \eta = 1.28 $
So:
$ \frac{1/\mu}{\sin 	heta_B} = 1.28 $
Step 3: Relation between $	heta_B$ and $\mu$
Since $	an 	heta_B = 	frac{1}{\mu}$,
$ \sin 	heta_B = \frac{	an 	heta_B}{\sqrt{1+	an^2 	heta_B}} = \frac{	frac{1}{\mu}}{\sqrt{1+	frac{1}{\mu^2}}} = \frac{1}{\sqrt{\mu^2+1}} $
Step 4: Substitute into ratio
$ \frac{\sin 	heta_c}{\sin 	heta_B} = \frac{	frac{1}{\mu}}{	frac{1}{\sqrt{\mu^2+1}}} = \frac{\sqrt{\mu^2+1}}{\mu} $
Given this equals 1.28:
$ \frac{\sqrt{\mu^2+1}}{\mu} = 1.28 $
Step 5: Solve for $\mu$
$ \sqrt{\mu^2+1} = 1.28 \mu $
Square both sides:
$ \mu^2 + 1 = 1.6384 \, \mu^2 $
$ 1 = (1.6384 - 1)\mu^2 $
$ 1 = 0.6384 \mu^2 $
$ \mu^2 = \frac{1}{0.6384} \approx 1.565 $
$ \mu \approx 1.25 $
Step 6: Find the relative index of the two media in the questionÃ¢ÂÂs terms
Here we defined $\mu = \frac{n_1}{n_2} \approx 1.25$.
But the
relative refractive index of the two media
usually means $\frac{n_2}{n_1}$ (rarer/denser), since that's how it appears in options (<1).
$ \frac{n_2}{n_1} = \frac{1}{\mu} \approx \frac{1}{1.25} = 0.8 $
Ã¢ÂÂ Final Answer:
$ \boxed{0.8} $
Correct Option: C

Answer

0.2

Answer

0.4

Answer

0.8

Answer

0.9

Ray Optics

A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $\theta iC and the Brewster's angle of incidence is \theta iB, such that sin\theta iC/sin\theta iB = \eta $ = 1.28. The relative refractive index of the two media is :

Choose the correct answer: