CBSE Class 12 Physics sample paper 2023-24: question 19

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Step-by-step solution

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1. For an equilateral prism $A=60^{\circ }$. Since $i=e$, the ray passes symmetrically and the deviation is minimum.
   $i=e=\frac{3}{4}A=\frac{3}{4}\times 60^{\circ }=45^{\circ }$.
   Using $i+e=A+{\delta }_m$: ${\delta }_m=2i-A=90^{\circ }-60^{\circ }=30^{\circ }$.

   1 mark

2. Refractive index: $n=\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}=\frac{\sin 45^{\circ }}{\sin 30^{\circ }}=\frac{1/\sqrt{2}}{1/2}=\sqrt{2}\approx 1.414$

   1 mark

Answer: Angle of deviation = 30° (minimum deviation); refractive index = √2 ≈ 1.41

Revise the chapter

• Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)