A parallel beam of light containing two wavelengths, ? 1 = 415 nm and ? 2 = 675
ID: 1376931 • Letter: A
Question
A parallel beam of light containing two wavelengths, ?1 = 415 nm and ?2 = 675 nm, enters the silicate flint glass of an equilateral prism as shown in Fig. 24-58 (? = 48
A parallel beam of light containing two wavelengths, ?1 = 415 nm and ?2 = 675 nm, enters the silicate flint glass of an equilateral prism as shown in Fig. 24-58 (? = 48°). At what angle, with respect to the normal, does each beam leave the prism?(Use the index of refraction 1.66 for violet light and 1.62 for red light) ?1 from the normal ° ?2 from the normal °Explanation / Answer
For red:
For the first interface,
Note that from Snell's law,
n1sin(t1) = n2sin(t2)
where
n1 = index of refraction of first medium = 1
t1 = angle of incidence = 48 degrees
n2 = index of refraction of second medium = 1.62
t2 = angle of refraction
Thus,
t2 = 27.30527562 degrees
Thus, for the seconf interface,
Note that from Snell's law,
n1sin(t1) = n2sin(t2)
where
n1 = index of refraction of first medium = 1.62
t1 = angle of incidence = 32.69472 degrees
n2 = index of refraction of second medium = 1
t2 = angle of refraction
Thus,
t2 = 61.05251393 degrees [final angle, red, ANSWER]
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For violet:
For the first interface,
Note that from Snell's law,
n1sin(t1) = n2sin(t2)
where
n1 = index of refraction of first medium = 1
t1 = angle of incidence = 48 degrees
n2 = index of refraction of second medium = 1.66
t2 = angle of refraction
Thus,
t2 = 26.59477889 degrees
For the second interface,
Note that from Snell's law,
n1sin(t1) = n2sin(t2)
where
n1 = index of refraction of first medium = 1.66
t1 = angle of incidence = 33.40522 degrees
n2 = index of refraction of second medium = 1.66
t2 = angle of refraction
Thus,
t2 = 33.40522 degrees [final angle, violet, ANSWER]
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