A ray of light impinges from air onto a block of ice (n = 1.309) at a 54.0 degre

Question

A ray of light impinges from air onto a block of ice (n = 1.309) at a 54.0 degree angle of incidence. Assuming that this angle remains the same, find the difference theta_2, ice - theta_2, water in the angles of refraction when the ice turns to water (n = 1.333) theta_2, ice - theta_2, water = The drawing shows a rectangular block of glass (n = 1.52) surrounded by liquid carbon disulfide (n = 1.63). A ray of light is incident on the glass at point A with a theta = 20.0 degree angle of incidence. At what angle of refraction does the ray leave the glass at point B? ___________________ degree A point source of light is submerged 3.0 m below the surface of a lake and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. What is the maximum radius that this circle could have?

Explanation / Answer

13)

We know that
According to Snell law, we have
     n1sin1 =n2sin2

1.63sin(20) = 1.52sin2,

sin2 = 1.63sin(20) / 1.52

Therefore the refraction angle at A is2 = 21.51o

The angle of incidence at B is 90 - 2=68.49o

      1.52sin(68.49) = 1.63sin(3)

sin 3 =1.52sin(68.49)/1.63

            3 = 60.17o

14)

1.333 sin A = sin90

A = 48.6

tan A = r/3
maximum radius that this circle , r = 3.402 m

Area = Pi*r2 = 36.377 m2

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