Question

A glass prism (u = 1.5) is dipped in water (u
= 4/3) as shown in figure. A light ray is incident
normally on the surface AB. It reaches the
surface BC after totally reflected, if
(1) sin 02 8/9
(2) 2/3 < sin < 8/9
(3) sin 0 < 2/3
(4) It is not possible​

Answer 1

Answer:

Sin ø >8/9

Explanation:

Applying smells law for the following, (1.5)sin(ø') =(4/3)sin(90)

Where ø' is critical angle where the light incident at the interface is just reflected

Sin(ø')=8/9

Therefore for any angle greater than arcsin(8/9) will be reflected of the surface due to refracting angle geater than 90 degrees

Hope this answer helped you

Answer 2

Answer:

Let the critical angle of the prism be C°, refractive index of prism be u_(p) and that of water be u_(w)

We know that,

$^{w}\mu_{p}=\frac{1}{sin\: C}\\ \Rightarrow \frac{\mu_{p}}{\mu_{w}}=\frac{1}{sin\: C}\\ \Rightarrow \frac{\mu_{w}}{\mu_{p}}=sin\: C\\ \Rightarrow \frac{\frac{4}{3}}{1.5}=sin\: C\\ \Rightarrow sin\: C=\frac{8}{9}$

Now,since theta>C (as the ray is )

=>sin theta>sin C ( internally reflected)

=>sin theta>8/9