By what angle (theta cornea) is a beam of light deviated as it passes from air t
ID: 1378713 • Letter: B
Question
By what angle (theta cornea) is a beam of light deviated as it passes from air to the cornea if the incident angle is 23.6 degrees? The refractive index of air is nair=1.00, the refractive index of the cornea is ncornea=1.38. Ignore further deviation of light as it passes from the cornea into the aqeous humour, ect.
b) The ability of your eyes to focus is impaired when you attempt to look around underwater. Recalculate your answer for the case in which the eye is submerged in water. (nwater=1.33)
Explanation / Answer
Note that from Snell's law,
n1sin(t1) = n2sin(t2)
where
n1 = index of refraction of first medium = 1
t1 = angle of incidence = 23.6 degrees
n2 = index of refraction of second medium = 1.38
t2 = angle of refraction
Thus,
t2 = 16.86 degrees [ANSWER]
******************
Note that from Snell's law,
n1sin(t1) = n2sin(t2)
where
n1 = index of refraction of first medium = 1.33
t1 = angle of incidence = 23.6 degrees
n2 = index of refraction of second medium = 1.38
t2 = angle of refraction
Thus,
t2 = 22.70 degrees [ANSWER]
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