CBSE Class 12 Physics 2026: question 33(a)

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Step-by-step solution

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1. (i)(I) Central maximum. At $\theta =0$, all secondary wavelets from the slit travel equal paths to the centre of the screen. They arrive in phase and add up, giving the bright central maximum.

   ½ mark

2. Minima. At angle $\theta$, the path difference between waves from the two edges of the slit is $a\sin \theta$. Suppose $a\sin \theta =\lambda$. Divide the slit into two equal halves. Each point in the upper half has a partner in the lower half with path difference $\lambda /2$, so every pair cancels and the intensity is zero. In general, minima occur at $a\sin \theta =n\lambda$ ($n=\pm 1,\pm 2,\dots$), i.e. at positions $y_n=n\lambda D/a$.

   1 mark

3. Secondary maxima. Suppose $a\sin \theta =\frac{3}{2}\lambda$. Divide the slit into three equal parts. The wavelets from two of the parts cancel, and the third part contributes, giving a bright fringe. In general, secondary maxima occur at $a\sin \theta =(n+\frac{1}{2})\lambda$.

   ½ mark

4. (i)(II) For the $n$th secondary maximum, the slit is divided into $(2n+1)$ equal parts. All but one part cancel in pairs, so only $\frac{1}{2n+1}$ of the slit contributes. As $n$ increases, this fraction shrinks, so successive maxima become weaker and weaker.

   1 mark

5. (ii) Differences (any two):
   1. Interference results from superposition of waves from two coherent sources (two slits). Diffraction results from superposition of secondary wavelets from different parts of the same wavefront.
   2. In interference, all bright fringes are of (nearly) equal intensity. In diffraction, the intensity of successive maxima falls rapidly.
   3. Interference fringes are of equal width. In diffraction, the central maximum is twice as wide as the other maxima.
   4. In double-slit interference, a maximum occurs at $\theta =\lambda /d$. In single-slit diffraction, a minimum occurs at $\theta =\lambda /a$.

   2 marks

Answer: Minima at $a\sin \theta =n\lambda$; secondary maxima at $a\sin \theta =(n+\frac{1}{2})\lambda$; the $n$th maximum gets only a $1/(2n+1)$ contribution of the slit, so it is weaker; differences: source of superposition, intensity distribution, fringe width

Revise the chapter

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