Question on: JAMB Physics - 2004

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 )