Calculate the angle of minimum deviation of a 6... - JAMB Physics 2017 Question
Calculate the angle of minimum deviation of a 60o prism of a refractive index [sin-10.75 = 49o]
A
38.00o
B
19.47o
C
16.25o
D
49o
correct option: a
For a refractive index () = (\frac{\sin\frac{1}{2} (A + D)}{\sin\frac{1}{2}A})
D = angle of minimum deviation
A = refractive angle of the prism
1.5 = (\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60})
1.5 = (\frac{\sin\frac{1}{2} (60 + D)}{\sin 30})
Sin (\frac{1}{2}) (60 + D) = 1.5 * Sin 30
Sin (\frac{1}{2}) (60 + D) = 0.75
(\frac{1}{2}) (60 + D) = Sin-1 (0.75)
but Sin-1 (0.75) = 49o
(\frac{1}{2}) (60 + D) = 49
60 + D = 2 * 49 = 98o
D = 98o - 60o
D = 38o
Please share this, thanks:
#JAMB #JAMB
Add your answer
No responses