CBSE Class 12 Physics 2023: question 33(a)

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Step-by-step solution

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1. (i)(1) Difference 1: An interference pattern has many equally spaced bright and dark fringes. A diffraction pattern has a central bright maximum twice as wide as the other maxima, and the secondary maxima get narrower and fainter.

   1 mark

2. Difference 2: In interference all bright fringes have (nearly) the same intensity. In diffraction the intensity of successive maxima falls rapidly on either side of the centre.

   (Also acceptable: interference comes from superposing two waves from two coherent narrow slits, while diffraction comes from superposing a continuous family of wavelets from every point of a single slit.)

   1 mark

3. (i)(2) Fringe width is $\beta =\frac{\lambda D}{d}$. So it depends on any two of:

   • the wavelength $\lambda$ of the light;
   • the distance $D$ of the screen from the slits;
   • the slit separation $d$.
   1 mark

4. (ii)(1) For the $n$th maximum, $d\sin \theta =n\lambda$.

   For the maximum next to the central one, $n=1$, so $\sin \theta =\frac{\lambda }{d}$.

   With $d=100\lambda$ and the small-angle approximation $\sin \theta \approx \theta$:

   $\theta =\frac{\lambda }{100\lambda }=\frac{1}{100}=0.01$ rad

   1 mark

5. (ii)(2) The distance between the two maxima on the screen is

   $\beta =D\theta =50\times 10^{-2}\text{m}\times \frac{1}{100}=5\times 10^{-3}\text{m}=5\text{mm}$

   1 mark

Answer: (ii)(1) 0.01 rad; (ii)(2) 5 mm

Revise the chapter

• Chapter: Wave Optics (notes, formulas, where marks are lost and every past question)
• Topic: Young's Double Slit Experiment