Please answer all parts and show work. Thank you. A very thin beam of white ligh

Question

Please answer all parts and show work. Thank you.

A very thin beam of white light is incident on a sheet of glass 1.35 am thick initially at an angle e 30.00 as shown here. a. If the index of refraction of the glass is 1.52 for red and 1.54 for violet, what is the angular spread as the light enters the sheet? b. After the light leaves the sheet of glass, a piece of paper is used to collect the rainbow. Depending on how one holds the paper a different sized rainbow can be measured, what is the smallest width of this rainbow? c. What is the largest angular spread that can be achieved inside this piece of glass by varying the initial incident angle) d. How wide is the rainbow formed on the back side of the glass from the situation described in part c? e. Now assume this same sheet of glass has been submerged in water with index of refraction of 1.33, answer parts C and D again with water on each side?

Explanation / Answer

a) Using Snell's Law we have n_a *sin(a) = n_b*sin(b)

so for red light we have

1*sin(30) = 1.52*sin(b)===> b = arcsin(sin(30)/1.52) = 19.21 deg

and for violet b = arcsin(sin(30)/1.54) = 18.97 deg

So the dispersion angle is 19.21 - 18.97 = 0.24 deg

Using trig we see that the violet beam strike the bottom of the glass at 1.35cm*tan(18.97) = 0.464cm
and the red beam strikes at 1.35cm*tan(19.21) = 0.470cm  

So the beam is 0.470 - 0.464 = 0.006cm wide

C) when the beams leave the glass they both will be refracted back at 30 deg

So the distance across the beam will be the separation distance times the cos(30)
= 0.006cm*cos(30) = 0.0052 cm

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