Rainbows from square drops. Suppose that, on some surreal world, raindrops had a

Question

Rainbows from square drops. Suppose that, on some surreal world, raindrops had a square cross section and always fell with one face horizontal. Figure 33-56 shows such a falling drop, with a white beam of sunlight incident at theta = 70.5o at point P. The part of the light that enters the drop then travels to point A, where some of it refracts out into the air and the rest reflects. That reflected light then travels to point B, where again some of the light refracts out into the air and the rest reflects. What is the difference in the angles of the red light (n = 1.3290) and the blue light (n = 1.3370) that emerge at (a) point A and (b) point B? (This angular difference in the light emerging at, say, point A would be the angular width of the rainbow you would see were you to intercept the light emerging there.)

Explanation / Answer

given

incident angle = 70.5

from the snells law

n1 sin1 = n2 sin2b

    where n2 = 1.3370

then 2b = sin^-1 ( 1/1.3370 * sin 70.5 ) = 44.83

n1 sin1 = n2 sin2r

where n2 = 1.3290

then 2r = sin^-1 ( 1/1.3290 * sin 70.5 ) = 45.17

for the refraction angle at the surface . thesae rays strike the second surface at complentary angles to those just caliculated . ta king this into consideration , we again use snells law to caliculate the second   refractions

using snells smillarly

then 3b= sin^-1 ( 1.3370 * sin (90- 2b)

then 3b= sin^-1 ( 1.3370 * sin (90-44.83 ) = 71.47

then 3r= sin^-1 ( 1.3290 * sin (90-45.17 ) = 69.54

which differs by 1.93

which gives the rain bow angular width

b) bothe of the refracteds rays emergs from the bottom side with same angle 70.5 with which they were incident on the top side does not alter this overall fact light comes into the block at the same angle that it emerges with from the opposite parallel side . there is no difference and thus theres is no rain bow in this case.

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