For total internal reflection, with i equal to the critical angle and r equal to 90 degrees, the relation simplifies. Using _i n_2 = sin i / sin r, what does the refractive index equal?

1. Start from $^i n_2 = \dfrac{\sin i}{\sin r}$. 2. At the critical angle, r = 90 degrees, so $\sin r = \sin 90^\circ = 1$. 3. Substitute: $^i n_2 = \dfrac{\sin i}{1} = \sin i$. 4. The chapter derives exactly this result. The trap cos i and tan i come from confusing the trig ratio; 1/sin i would only appear if the index were inverted. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 6 "Refraction of Light", p.87_