The Figure shows a light beam hitting a glass semi-circle of index of refraction
ID: 1263996 • Letter: T
Question
The Figure shows a light beam hitting a glass semi-circle of index of refraction n1= 1.41 a distance d = 4.0 cm below the center. The radius of the semi-circle is R = 8.0cm and it is surrounded by air (n = 1).
a) What is the angle of refraction of the beam into the glass?
The light beam will obviously move inside the glass and hit the curved interface between the glass and air.
b) What is the angle of incidence that the beam will make with this curved interface?
c) What is the critical angle between this glass-air interface?
d) At what angle relative to the normal would the beam exit the glass?
Explanation / Answer
as per snells law
n1 sin theta 1 = n2 sin theta 2
if incidence angle is known the refracted angle can be found as n1 and n2 are known
for air n =1
Angle of incidence on curved surface = anle of refraction at the plane surface
critical angle :-
n2 sin theta = n1 sin 90
sin theta c = 1/1.41
theta = 45.17 degrees
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