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An aquarium filled with water has flat glass sides whose index of refraction is 1.48. A beam of light from outside the aquarium...Asked by ####Nadine####
An aquarium filled with water has flat glass sides whose index of refraction is 1.48. A beam of light from outside the aquarium strikes the glass at a 43.5º angle to the perpendicular . The angle this light ray enters the glass is 27.7º, what is the angle of this light ray when it enters the water?
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Answered by
Elena
The angle in the glass ( r ) related to the angle of incidence (i) as
sin i/sin r = n2/n1,
sin r = n1•sin i/n2 = 1•sin 43.5/1.48 = =0.465.
r = arcsin 0.465= 27.7degr.
Because the surfaces are parallel, the refraction angle from the first surface is the incident angle at the second surface.
sin r/sin φ =n3/n2,
sin φ = n2•sin r/n3 =1.48•0.465/1.33 =0.517,
φ = arcsin 0.517 =31.16 degr.
sin i/sin r = n2/n1,
sin r = n1•sin i/n2 = 1•sin 43.5/1.48 = =0.465.
r = arcsin 0.465= 27.7degr.
Because the surfaces are parallel, the refraction angle from the first surface is the incident angle at the second surface.
sin r/sin φ =n3/n2,
sin φ = n2•sin r/n3 =1.48•0.465/1.33 =0.517,
φ = arcsin 0.517 =31.16 degr.
Answered by
####Nadine####
Thank you, I did that but I did not round until the end so I had 31.06º which was listed as incorrect
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