A ray of light travelling in glass $\left( \mu = \frac{3}{2} ight)\ $is incident on a horizontal glass air surface at the critical angle $ heta_{c}.$ If thin layer of water $\left( \mu = \frac{4}{3} ight)\ $is now poured on the glass air surface, the angle at which the ray emerges into air the water-air surface is [image:9_image54.emf]

Question

A ray of light travelling in glass $\left( \mu = \frac{3}{2} ight)\ $is incident on a horizontal glass air surface at the critical angle $	heta_{c}.$ If thin layer of water $\left( \mu = \frac{4}{3} ight)\ $is now poured on the glass air surface, the angle at which the ray emerges into air the water-air surface is [image:9_image54.emf]

Dr. Abubacker · Accepted Answer

(c) $\mu_{g}\sin{	heta_{c} = \mu_{1}\sin{90{^\circ}}}$
Or ${\ \mu}_{g}\sin{	heta_{c} = 1}$
When water is poured,
$\mu_{w}{sin\ r = \mu_{s}\ sin\ 	heta_{c}\ or\ \ \mu_{w}\ sin\ r = 1}\ $
Again, $\mu_{a}\sin{	heta = \mu_{w}\sin r}$
Or ${\ \mu}_{a}\sin{	heta = 1}$
Or $\ \sin{	heta = 1\ or\ \ 	heta = 90{^\circ}}$

Explanation

Reviewed by Dr. Abubacker

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