REFRACTION OF LIGHT AND REFRACTIVE INDEX
12.A ray of light incident on a transparent block at an angle of incidence 60degrees.If the refractive index of the block is 1.732, the angle of deviation of the refracted ray is--------------

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Solution

$A n g l e o f i n c i d e n c e I = 60 d e g r e e a n g l e o f r e f r a c t i o n = r r e f r a c t i v e i n d e x = 1.732 r e f r a c t i v e i n d e x = \frac{S i n r}{S i n i} 1.732 = \frac{\sin 60}{\sin r} \sin r = \frac{\sqrt{3}}{2 \times 1.732} = \frac{1}{2} r = 30 d e g r e e d e v i a t i o n = a n g l e o f i n c i d n e c e - a n g l e o f r e f r a c t i o n = 60 - 30 = 30 d e g r e e$

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REFRACTION OF LIGHT AND REFRACTIVE INDEX12.A ray of light incident on a transparent block at an angle of incidence 60degrees.If the refractive index of the block is 1.732, the angle of deviation of the refracted ray is--------------

Angle of incidence I = 60 degreeangle of refraction = rrefractive index= 1.732refractive index= Sin rSin i1.732 = sin 60sin rsin r = 32×1.732 =12r = 30 degreedeviation = angle of incidnece - angle of refraction =60-30 = 30 degree

REFRACTION OF LIGHT AND REFRACTIVE INDEX
12.A ray of light incident on a transparent block at an angle of incidence 60degrees.If the refractive index of the block is 1.732, the angle of deviation of the refracted ray is--------------