WAEC - Physics (1991 - No. 25)
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)
10.5o
19.5o
21.1o
38.9o
40.5o
Тайлбар
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