(a) Given an interface between clear tissue (i.e. water) n_t = 1.33 and glass (n

Question

(a) Given an interface between clear tissue (i.e. water) n_t = 1.33 and glass (n_g = 1.5), compute the transmission angle for a beam incident in the water at 45 degree. If the transmitted beam is reversed, so that it impinges on the interface from the glass side, calculate what the transmitted angle will be through the tissue. Does this make sense? Explain. (b) Using MATLAB, or some other software, plot (R_perp = r_perp^2) and (R_p = r_p^2) as a function of the input angle, theta for a glass (n_t = 1.5) and air (n_t = 1.0) interface. Explain what it means.

Explanation / Answer

a. ffor the given interface

ni = 1.33 [ refractgive index of incident medium, water]

nr = 1.5 [ refractive index of refractive medium, glass]

so given angle of incidence thetai = 45 deg

then angle of transmission ( refraction ) = tehtar

from snells law

sin(theta)i/sin(thetar) = nr/ni

then sin(45)/sin(thetar) = 1.5/1.33

thetar = 38.82 deg

for the reverse light

thetai = 38.2 deg

from snells law

sin(thetai)/sin(thetar) = 1.33/1.5

thetar = 45 deg

so the light ray retraces its path as it is

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