Name: Total Internal Reflection From Prism
Uploaded: 2019-07-19
Duration: 1 min 56 s
Description: You must have seen reflectors on a bicycle. Why are they attached to the mudguards? What use do they have? Let’s watch this video on reflectors to find out.

Question

A ray of light incident normally on one face of a right angled isosceles prism is found to be totally reflected Fig. $13.77$ . What is the minimum value of the refractive index of the material of the prism? When the prism is immersed in water, trace the path of the emergent ray for the same incident ray, indicating values of all the angles. $(μ_{w} = \frac{4}{3})$ .

Accepted Answer

ABC is the section of the prism, B is a right angle. A and C are equal angles,i.e, $A = C = 45 °$ . The ray PQ is normally incident on the face AB. Hence, it is normally refracted and the ray QR strikes the face AC at angle of incidence. $45 °$ . It is given that the ray does not undergo refraction but is totally reflected at the face AC. This given a maximum value for the critical angle at $45 °$ .

$\sin C = \sin 45 ° = \frac{1}{\sqrt{2}} in the limit$

$Since, μ = \frac{1}{\sin C} = \frac{1}{\sqrt{2}}$

Or, $μ = \frac{1}{\sin 45 °}, μ_{m i n} = \sqrt{2} = 1.414$

Now, when the prism is immersed in water the critical angle for the glass-water interface is given by

$\sin C_{1} = \frac{4 / 3}{\sqrt{2}} = \frac{4}{3 \sqrt{2}} = 70.53 °$

The angle of incidence at R continues to be $45 °$ And since $45 ° < 70.53 °$

There is refraction taking place now and the refracted rat is RS. The angle of refraction is given by

$μ_{g} \sin i = μ_{w} \sin r$

${}_{g}{μ_{w} = \frac{\sin i}{\sin r}}$

$\sqrt{2} \sin 45 ° = \frac{4}{3} \sin r \sin r = \frac{3 \sqrt{2}}{4} \times \frac{1}{\sqrt{2}}$

$r = \sin^{- 1} \frac{3}{4}$

Important Questions on Refraction of Light

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