Question 235: Ray Optics

Question

Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC . At an angle of incidence A when light is travelling from
denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90o. Angle A is given by :

Accepted Answer

Refractive index,
n
12
= $${{{n_D}} \over {{n_R}}}$$ = $${1 \over {\sin {	heta _c}}}$$
$$ \Rightarrow $$ $$\,\,\,$$ sin$$	heta $$
c
= $${{{n_R}} \over {{n_D}}}$$ . . . . . (1)
From Snell's law,
n
D
sinA = n
R
sin r
$${{{n_R}} \over {{n_D}}}$$ = $${{\sin A} \over {\sin \left( {{{90}^o} - A} ight)}}$$    [as    r = 90
o
$$-$$ A]
$$ \Rightarrow $$$$\,\,\,$$ $${{{n_R}} \over {{n_D}}} = 	an A$$
$$	herefore\,\,\,$$ From (1) we get,
tan A = sin $$	heta $$
c
$$ \Rightarrow $$$$\,\,\,$$ A = tan
$$-$$1
(sin $$	heta $$
c
)

Answer

$${1 \over {{{\cos }^{ - 1}}\left( {\sin {	heta _C}} ight)}}$$

Answer

$${1 \over {{{	an }^{ - 1}}\left( {\sin {	heta _C}} ight)}}$$

Answer

$${\cos ^{ - 1}}\,\left( {\sin {	heta _C}} ight)$$

Answer

$${	an ^{ - 1}}\,\left( {\sin {	heta _C}} ight)$$

Ray Optics

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