a ray of light which is incident in air on the surface of a glass block is deviated through 15 degre. the angle of incidence in air is 60 degre . what is the angle of refraction in glass also calculate the critical angle for glass in air interface?

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Solution

Angle of incidence = 60 $°$ , Angle of deviation = 15 $°$ , angle of refraction = 60-15 = 45 $°$
$μ_{g} = \frac{\sin i}{\sin r} = \frac{\sin 60}{\sin 45} = 1.225 \sin c = \frac{1}{μ_{g}} c = \sin^{- 1} (1 / 1.225) = 55 ° 5'$

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a ray of light which is incident in air on the surface of a glass block is deviated through 15 degre. the angle of incidence in air is 60 degre . what is the angle of refraction in glass also calculate the critical angle for glass in air interface?

Angle of incidence = 60°, Angle of deviation = 15°, angle of refraction = 60-15 = 45 ° μg=sin isin r= sin60sin45=1.225sin c =1μgc=sin-1(1/1.225)=55°5'

Angle of incidence = 60 $°$ , Angle of deviation = 15 $°$ , angle of refraction = 60-15 = 45 $°$
$μ_{g} = \frac{\sin i}{\sin r} = \frac{\sin 60}{\sin 45} = 1.225 \sin c = \frac{1}{μ_{g}} c = \sin^{- 1} (1 / 1.225) = 55 ° 5'$