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)
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option A
Explanation:
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
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