A ray of light is travelling from glass to air. (Refractive index of glass $= 1.5$ ). The angle of incidence is $50^{o}$ .The deviation of the ray is
$0^{o}$
80 $^{o}$
$s i n^{- 1} [\frac{s i n 50^{o}}{1.5}] - 50^{o}$
$50^{o} - s i n^{- 1} [\frac{s i n 50^{o}}{1.5}]$

Question

A ray of light is travelling from glass to air. (Refractive index of glass $= 1.5$ ). The angle of incidence is $50^{o}$ .The deviation of the ray is
$0^{o}$
80 $^{o}$
$s i n^{- 1} [\frac{s i n 50^{o}}{1.5}] - 50^{o}$
$50^{o} - s i n^{- 1} [\frac{s i n 50^{o}}{1.5}]$

Toppr Admin · Accepted Answer

A-x3BC-g-1-5-x2234-1-5-cosecC-xA0-Or- C-420-xA0-Critical angle for glass -420- Hence a ray of light incident at 500 in glass medium undergoes total internal reflection- -x3B4- denotes the deviation of the ray-x3B4-180o-x2212-50o-50o- or -x3B4-80o-

A ray of light is travelling from glass to air. (Refractive index of glass =1.5). The angle of incidence is 50o.The deviation of the ray is0o80osin−1[sin50o1.5]−50o50o−sin−1[sin50o1.5]

aμg=1.5∴1.5=cosecC Or, C=420. Critical angle for glass =420. Hence a ray of light incident at 500 in glass medium undergoes total internal reflection. δ denotes the deviation of the ray.δ=180o−(50o+50o) or δ=80o.

A ray of light is travelling from glass to air. (Refractive index of glass $= 1.5$ ). The angle of incidence is $50^{o}$ .The deviation of the ray is
$0^{o}$
80 $^{o}$
$s i n^{- 1} [\frac{s i n 50^{o}}{1.5}] - 50^{o}$
$50^{o} - s i n^{- 1} [\frac{s i n 50^{o}}{1.5}]$

$^{a} μ_{g} = 1.5$

$∴ 1.5 = c o s e c C$

Or, $C = 42^{0}$ .

Critical angle for glass $= 42^{0}$ . Hence a ray of light incident at $50^{0}$ in glass medium undergoes total internal reflection. $δ$ denotes the deviation of the ray.

$δ = 180^{o} - (50^{o} + 50^{o})$ or $δ = 80^{o}$ .