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light strikes a 5.0-cm thick sheet of glass at an angle of incidence in air of 50°. the sheet has parallel faces and the glass has an index of refraction 1.50. a. what is the angle of refraction in the glass? b. after traveling through the glass the light re-emerges into the air. what is the final angle of refraction in air? c. as it leaves the glass, by what distance is the path of the ray is displaced from what it was before entering the glass?

Sagot :

The angle of refraction in the glass is 30°.

Calculation

i= 50°

u= 1.50

u1sini=u2sinr

⇒sinr =[tex]\frac{sin 50}{1.5\\}[/tex]

=0.50=1/2

so angle r = [tex]sin^{-1}[/tex](0.5) =30°

Then we have to calculate r final.

[tex]u_{glass}[/tex]sini= [tex]u_{air}[/tex] sinr

⇒ r = [tex]sin^{-1}[/tex] (1.5 x [tex]\frac{\sqrt{3} }{2}[/tex])

[tex]sin^{-1}[/tex](1.29) = 0.0225

Then the displacement can be calculated as follows:

T= thickness = 5cm

s= t (1- 1/u)

s= 2cm

Thus, all the values have been found out.

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