CBSE Class 12 Physics 2025: question 33(b)

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Step-by-step solution

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1. (i) Let the two coherent waves at a point be $y_1=a\cos \omega t$ and $y_2=a\cos (\omega t+\varphi )$.

   By the principle of superposition:
   $y=y_1+y_2=a[\cos \omega t+\cos (\omega t+\varphi )]=2a\cos \frac{\varphi }{2}\cos \left(\omega t+\frac{\varphi }{2}\right)$.

   1 mark

2. So the resultant amplitude is $A=2a\cos \frac{\varphi }{2}$.

   Since $I\propto A^2$: $I=k4a^2{\cos }^2\frac{\varphi }{2}$. With $I_0=k{a}^2$:

   $$I=4I_0{\cos }^2\frac{\varphi }{2}$$

   1 mark

3. Maximum: $I_{max}=4I_0$, when ${\cos }^2\frac{\varphi }{2}=1$, i.e. $\varphi =0,2\pi ,4\pi ,\dots$

   Minimum: $I_{min}=0$, when ${\cos }^2\frac{\varphi }{2}=0$, i.e. $\varphi =\pi ,3\pi ,\dots$

   1 mark

4. (ii)(I) Data: $a=3\times 10^{-3}$ m, $D=1.5$ m, $\lambda =600\times 10^{-9}$ m.

   The $n$th minimum is at $y_n=\frac{n\lambda D}{a}$. For $n=1$:

   $y_1=\frac{600\times 10^{-9}\times 1.5}{3\times 10^{-3}}=3\times 10^{-4}$ m $=0.3$ mm.

   1 mark

5. (II) Secondary maxima are at $y=\frac{(2n+1)\lambda D}{2a}$. For the second order maximum, $n=2$:

   $y_2=\frac{5\lambda D}{2a}=\frac{5\times 600\times 10^{-9}\times 1.5}{2\times 3\times 10^{-3}}=7.5\times 10^{-4}$ m $=0.75$ mm.

   1 mark

Answer: (i) $I=4I_0{\cos }^2(\varphi /2)$; $I_{max}=4I_0$, $I_{min}=0$; (ii)(I) $3\times 10^{-4}$ m; (II) $7.5\times 10^{-4}$ m

Revise the chapter

• Chapter: Wave Optics (notes, formulas, where marks are lost and every past question)
• Topic: Single Slit Diffraction