Calculate the angle of minimum deviation for a ray which is refracted through an equiangular prism of refractive index 1.4
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
From the relation refractive index = Sin \( \left(\frac{A+Dm}{2}\right) \)
Where A = refractive angle = 600, Sin\( \left(\frac{A}{2}\right) \)
\( \implies 1.4 = \frac{\text{Sin}\left(\frac{60^0 + Dm}{2}\right)}{\text{Sin}\left(\frac{60}{2}\right)} \\
= \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{\text{Sin}30^0} \\
= \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{0.5}\\
\text{Therefore} 1.4 \times 0.5 = \text{Sin}\left(\frac{60 + Dm}{2}\right) \\
\text{Therefore} \left(\frac{60 + D_m}{2}\right)= \text{Sin}^{-1} 0.7000 \\
= 44^0 12^1 \\
60^0 + D_m = 88^0 24^1 \\
D_m = 88^0 24^1 - 60^0 \\
= 28^0 24 \\
29^0 \)
From the relation refractive index = Sin \( \left(\frac{A+Dm}{2}\right) \)
Where A = refractive angle = 600, Sin\( \left(\frac{A}{2}\right) \)
\( \implies 1.4 = \frac{\text{Sin}\left(\frac{60^0 + Dm}{2}\right)}{\text{Sin}\left(\frac{60}{2}\right)} \\
= \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{\text{Sin}30^0} \\
= \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{0.5}\\
\text{Therefore} 1.4 \times 0.5 = \text{Sin}\left(\frac{60 + Dm}{2}\right) \\
\text{Therefore} \left(\frac{60 + D_m}{2}\right)= \text{Sin}^{-1} 0.7000 \\
= 44^0 12^1 \\
60^0 + D_m = 88^0 24^1 \\
D_m = 88^0 24^1 - 60^0 \\
= 28^0 24 \\
29^0 \)