(a) A light beam in glass (n = 1.5) reaches an air-glass interference, at an ang

Question

(a) A light beam in glass (n = 1.5) reaches an air-glass interference, at an angle of 60 degrees from the surface. What is the angle of the refracted light beam from the surface in air? Note: sin(30) = 0.5, sin(45) = 0.71, sin(60) = 0.87 (all angles are in degrees)

Possible answers:

There is no refracted ray

arcsin(1.5*0.71)    

arcsin(0.5/1.5)

arcsin(0.71/1.5)

arcsin(1.5*0.87)

arcsin(0.87/1.5)

arcsin(1.5*0.5)

(b) A light beam in glass (n = 1.5) reaches an air-glass interference, at an angle of 30 degrees from the surface. What is the angle of the refracted light beam from the surface in air? <_>

Possible answers:

arcsin(0.71/1.5)

arcsin(1.5*0.87)     

arcsin(1.5*0.5)

arcsin(0.5/1.5)

There is no refracted ray

arcsin(1.5*0.71)

arcsin(0.87/1.5)

Explanation / Answer

here,

a)

refractive index of glass , n1 = 1.5

the angle of incidence , i = 30 degree

the angle of refraction , theta = arcsin(n1 * sin(theta1))

theta = arcsin(1.5 * 0.5)

the angle of the refracted light beam from the surface in air is arcsin(1.5 * 0.5) or the angle of refraction is 48.59 degree

b)

refractive index of glass , n1 = 1.5

the angle of incidence , i = 30 degree

the angle of refraction , theta = arcsin(n1 * sin(theta1))

theta = arcsin(1.5 * 0.87)

that is not possible

therefore there is no refracted ray

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