A parallel-sided plate of glass having a refractive index of 1.60 is in contact
ID: 1392020 • Letter: A
Question
A parallel-sided plate of glass having a refractive index of 1.60 is in contact with the surface of water in a tank. A ray coming from above makes an angle of incidence of 32.0? with the top surface of the glass. What angle does this ray make with the normal in the water? A parallel-sided plate of glass having a refractive index of 1.60 is in contact with the surface of water in a tank. A ray coming from above makes an angle of incidence of 32.0? with the top surface of the glass. What angle does this ray make with the normal in the water? A parallel-sided plate of glass having a refractive index of 1.60 is in contact with the surface of water in a tank. A ray coming from above makes an angle of incidence of 32.0? with the top surface of the glass. What angle does this ray make with the normal in the water?Explanation / Answer
Sin (Theta1) / Sin(Theta 2) = n_glass / n_air
n_glass = 1.60
n_air = 1
theta1 = 32 degree
Sin(32) / Sin(Theta2) = 1.6/1
1*Sin(32)/1.6 = Sin(theta2)
Theta2 = 19.34 degree
then
n_glass = 1.6
n_water = 1.33
Sin(Theta2) / Sin(Theta3) = n_water / n_glass
Sin(19.34deg) / Sin(Theta3) = 1.33 / 1.6
Sin(Theta3) = 1.6/1.33 * sin(19.34deg)
Sin(Theta3) = 0.3984
Theta3 = sin-1 (0.3984) = 23.48degree
so the angle this ray make with the normal in the water is 23.48 degree
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