Skip to content
PhysicsNEET 2024Total internal reflection and prisms

Ray Optics and Optical Instruments

A light ray enters through a right-angled prism at point PP with the angle of incidence 30∘30^\circ as shown in the figure. It travels through the prism parallel to its base BCBC and emerges along the face ACAC. The refractive index of the prism is:

Question 6 as printed in the NEET 2024 booklet

  1. A.52\dfrac{\sqrt{5}}{2}✓
  2. B.34\dfrac{\sqrt{3}}{4}
  3. C.32\dfrac{\sqrt{3}}{2}
  4. D.54\dfrac{\sqrt{5}}{4}

Solution

Let the prism angles at B and C be BB and CC with B+C=90∘B + C = 90^\circ. The ray inside is parallel to BC, so its angle of refraction at AB is r=90∘−Br = 90^\circ - B and its angle of incidence at AC is 90∘−C=B90^\circ - C = B. Emerging along AC means BB is the critical angle: nsin⁡B=1n\sin B = 1. At P, sin⁡30∘=nsin⁡r=ncos⁡B\sin 30^\circ = n\sin r = n\cos B, so ncos⁡B=12n\cos B = \tfrac12. Squaring and adding: n2=1+14⇒n=52n^2 = 1 + \tfrac14 \Rightarrow n = \dfrac{\sqrt5}{2}.

Save for spaced revision (free account)

More from this chapter in past papers

From Physics › Ray Optics and Optical Instruments › Total internal reflection and prisms — lectures, notes and every past question on it.

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram