Question

During an optical experiment, a glass container is placed in an ambient gas with index of refraction ngas= 1.75. Liquid 1 floats on top of liquid 2. A light ray enters at an incidence angle of 63 degrees into liquid 1. The liquids have indices of refraction n1= 1.80 and n2= 1.70.

A) What angle does the ray inside liquid 1 make with the normal?

B) What angle does the ray inside liquid 2 make with the normal?

C) If the ray inside liquid 2 experiences tota internal reflection at the bottom of the glass, what is the minimum index of refraction of the glass?

D) The reflected ray then reaches the right side of the container. At what angle does it refract inside the glass?

Answer 1

Note that from Snell's law,          
          
n1sin(t1) = n2sin(t2)          
          
where          
          
n1 = index of refraction of first medium =    1.75      
t1 = angle of incidence =    63   degrees  
n2 = index of refraction of second medium =    1.8      
t2 = angle of refraction          
Thus,          
          
t2 =    60.02647428   degrees   [ANSWER]

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Note that from Snell's law,          
          
n1sin(t1) = n2sin(t2)          
          
where          
          
n1 = index of refraction of first medium =    1.8      
t1 = angle of incidence =    60.0264   degrees  
n2 = index of refraction of second medium =    1.7      
t2 = angle of refraction
          
Thus,          
          
t2 =    66.52182703   degrees   [ANSWER]

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If the critical angle is 66.521 degrees, then

sin 66.521 = nglass / 1.70

Thus,

nglass = 1.56   [ANSWER]

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The reflected ray will have an angle of 90 - 66.521 = 23.479 degrees.

Thus,

Note that from Snell's law,          
          
n1sin(t1) = n2sin(t2)          
          
where          
          
n1 = index of refraction of first medium =    1.7      
t1 = angle of incidence =    23.479   degrees  
n2 = index of refraction of second medium =    1.559      
t2 = angle of refraction
          
Thus,          
          
t2 =    25.7500739   degrees   [ANSWER]