Q. Three right angled prisms of refractive indices $n_{1},\ n_{2}$ and $n_{3}$ are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then
A) $n_{1} = n_{2} = n_{3}$
B) $n_{1} = n_{2} \neq n_{3}$
C) $1 + n_{1} = n_{2} + n_{3}$
D) $1 + n_{2}^{2} = n_{1}^{2} = n_{3}^{2}$
Explanation
(d) At $B$
$\sin i = n_{1}\sin r_{1} \Rightarrow \sin^{2}i = n_{1}^{2}\sin^{2}r_{1}$ …(i)
At $C$
$n_{1}\sin{(90 - r_{1})} = n_{2}\sin r_{2} \Rightarrow n_{2}^{2}\sin^{2}r_{2} = n_{1}^{2}\cos^{2}r_{1}$ …(i)
At $D$
$n_{2}\sin{(90 - r_{2})} = n_{3}\sin r_{3} \Rightarrow n_{2}^{2}\cos^{2}r_{2} = n_{3}^{2}\sin^{2}r_{3}$ …(iii)
At $E$
$n_{3}\sin{(90 - r_{3})} = (1)\sin{(90 - i)} \Rightarrow \cos^{2}{i =}n_{3}^{2}\cos^{2}r_{3}$ …(iv)
Adding (i), (ii), (iii), and (iv) we get $1 + n_{2}^{2} = n_{1}^{2} + n_{3}^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.