CBSE Class 12 Physics 2023: question 26(a)

Hints

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Step-by-step solution

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1. Construction (diagram described): Let $P{P}^{\prime }$ be the interface. A plane wavefront $AB$ is incident at angle $i$, with point $A$ on the surface. Let $\tau$ be the time the wavefront takes to travel from $B$ to $C$ on the surface, so $BC=v_1\tau$. With centre $A$ and radius $AE=v_2\tau$, draw a secondary wavelet (sphere) in medium 2. Since ${\mu }_2>{\mu }_1$, $v_2<v_1$ and so $AE<BC$. Draw the tangent plane $CE$ from $C$ to this sphere. $CE$ is the refracted wavefront and it bends towards the normal. Show the incident and refracted rays perpendicular to the wavefronts.

   1 mark

2. In right triangle $ABC$ (right-angled at $B$): $\sin i=\frac{BC}{AC}=\frac{v_1\tau }{AC}$.

   ½ mark

3. In right triangle $AEC$ (right-angled at $E$): $\sin r=\frac{AE}{AC}=\frac{v_2\tau }{AC}$, where $r$ is the angle of refraction.

   ½ mark

4. Dividing the two results: $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$. Using $v_1=c/{\mu }_1$ and $v_2=c/{\mu }_2$, this becomes $\frac{\sin i}{\sin r}=\frac{{\mu }_2}{{\mu }_1}$.

   ½ mark

5. Hence ${\mu }_1\sin i={\mu }_2\sin r$, which is Snell's law of refraction.

   ½ mark

Answer: Refracted wavefront CE (tangent from C to the wavelet of radius v₂τ); sin i/sin r = v₁/v₂ = μ₂/μ₁ ⇒ μ₁ sin i = μ₂ sin r

Revise the chapter

• Chapter: Wave Optics (notes, formulas, where marks are lost and every past question)
• Topic: Huygens Principle and Wavefronts