a) What angle would the beam that leaves the glass surface (after being reflecte

Question

a) What angle would the beam that leaves the glass surface (after being reflected from the silver backing) make with the normal to the glass surface? Provide the answer in degrees.

b)  How far is the point at which the beam leaves the glass surface (after being reflected from the silver backing) from the point at which the beam entered the glass? Provide the answer in mm.

A beam of light makes an angle of 47.0 with the normal of a mirror made of 4.00-mm-thick glass silvered on the back. The index of refraction of the glass is 1.28. a) What angle would the beam that leaves the glass surface (after being reflected from the silver backing) make with the normal to the glass surface? Provide the answer in degrees. b) How far is the point at which the beam leaves the glass surface (after being reflected from the silver backing) from the point at which the beam entered the glass? Provide the answer in mm.

Explanation / Answer

Note that, even with refraction in the middle, for plane mirrors like this, light will still escape as an ordinary reflected angle equal to the incident angle, (t_incident = t_reflected)

t2 = 47.0 degrees.   [ANSWER, PART A]

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Note that from Snell's law,      
      
n1sin(t1) = n2sin(t2)      
      
where      
      
n1 = index of refraction of first medium =    1  
t1 = angle of incidence =    47   degrees
n2 = index of refraction of second medium =    1.28  
t2 = angle of refraction
      
Thus,      
      
t2 =    34.84582092   degrees

Now, the distance d is given by

d = 2L tan(t2)

where L = the thickness = 4.0 mm, and the factor 2 is because it does this twice, back and forth.

Thus,

d = 5.57 mm   [ANSWER, PART B]

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