Question on: WAEC Physics - 1991
A ray of light is incident at an angle of 30o on a glass prism of refractive index 1.5. Calculate the angle through which the ray is minimally deviated in the prism. (The medium surrounding the prism is air)
A
10.5o
B
19.5o
C
21.1o
D
38.9o
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Correct Option: A
n = \(\frac{sin i}{sin r}\)
sin r = \(\frac{30^o}{n}\)
sin r = \(\frac{0.5}{1.5}\)
sin r = 0.3333
r = 19.5o; Deviation 30 - 19.5 = 10.5o
sin r = \(\frac{30^o}{n}\)
sin r = \(\frac{0.5}{1.5}\)
sin r = 0.3333
r = 19.5o; Deviation 30 - 19.5 = 10.5o
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