Question

In the adjacent figure, find the refractive index of denser medium if refractive index of rarer medium

is 1?​

Answer 1

Answer:

Given:

A Refraction setup has been provided , such that the incident ray falls at an angle of 60° and refracted ray moves at an angle of 90° to the normal

To find:

Refractive Index of denser medium.

Concept:

This is a critical case when the refracted ray grazes the surface after Refraction. A slight increase in Incidence Angle will result in Total Internal Reflection.

Calculation:

Applying Snell's Law :

$\mu1 \times \sin(i) = \mu2 \times \sin(r)$

$= > \mu1 \times \sin(60 \degree) = 1 \times \sin(90 \degree)$

$= > \mu1 = \dfrac{1}{ \sin(60 \degree) }$

$= > \mu1 = \dfrac{1}{ (\frac{ \sqrt{3} }{2} )}$

$= > \mu1 = \dfrac{2}{ \sqrt{3} }$

So final answer :

$\boxed{ \blue{ \bold{ \huge{ \mu1 = \dfrac{2}{ \sqrt{3} } }}}}$

Answer 2

Solution :

⏭ Given:

✏ Angle of incident = 60°

✏ Angle of refraction = 90°

✏ Refractive index of rarer medium = 1

⏭ To Find:

✏ Refractive index of denser medium

⏭ Concept:

✏ This question is completely based on concept of critical angle.

✏ Critical angle is the angle made in denser medium for which the angle of refraction in rarer medium is 90°. When angle in denser medium is more then crirical angle the light ray reflects back in denser medium following the laws of reflection and the interface behaves like a perfectly reflecting mirror.

⏭ Calculation:

✏ As per snell's law...

$\implies \sf \: \red{ \mu_{1} \sin \theta_{1} = \mu_{2} \sin{ \theta_{2}}} \\ \\ \implies \sf \: \mu_{1} \sin60 \degree = 1 \times \sin90 \degree \\ \\ \implies \sf \: \mu_{1} \times \frac{ \sqrt{3}}{2} = 1 \\ \\ \implies \: \underline{ \boxed{ \bold{ \sf{ \pink{ \large{ \mu_{1} = \frac{2}{ \sqrt{3}}}}}}} } \: \green{ \bigstar}$

⏭ Additional information:

• Refractive index is unitless as well as dimensionless quantity.