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missmontgomery21
23.07.2019 •
Physics
The critical angle for total internal reflection at a liquid-air interface is 42.0 ∘. if a ray of light traveling in the liquid has an angle of incidence at the interface of 30.5 ∘, what angle does the refracted ray in the air make with the normal? if a ray of light traveling in air has an angle of incidence at the interface of 30.5 ∘, what angle does the refracted ray in the liquid make with the normal?
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Ответ:
sin i / sin r = u (refractive index of air with respect to liquid)
so sin 42 / sin 90 = u since sin 90 =1 therefore, u of liquid w.r.t. air = sin 42 deg.
Case 1 since /_ i = 30.5 deg sin 30.5/ sin r = u (which is sin 42 deg) so sin r = sin 31 / u angle r = sin inverse of (sin 30.5 /sin 42) angle r = 49.33
Case 2 the ray travels from air to liquid
So the refractive index of liquid wrt air = 1/u
angle i = 30.5 angle r =? sin 30.5 / sin r = 1/u = 1/sin 42 so, sin r = sin 30.5 x sin 40 angle r = arc sin (sin 30.5 x sin 40)angle r = 19.04 degrees
Ответ:
Fruits are sources of many essential nutrients that are under consumed, including potassium, dietary fiber, vitamin C, and folate (folic acid).
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