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Light Waves — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsLight Waves5 Marks
Question
In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to,
  1. Half the maximum.
  2. One fourth of the maximum.
✓

Answer

  1. When intensity is half the maximum $\frac{\text{I}}{\text{I}_\text{max}}=\frac{1}{2}$

$\Rightarrow\frac{4\text{a}^2\cos^2\big(\frac{\phi}{2}\big)}{4\text{a}^2}=\frac{1}{2}$

$\Rightarrow\cos^2\Big(\frac{\phi}{2}\Big)=\frac{1}{2}\Rightarrow\cos\Big(\frac{\phi}{2}\Big)=\frac{1}{\sqrt{2}}$

$\Rightarrow\frac{\phi}{2}=\frac{\pi}{4}\Rightarrow\phi=\frac{\pi}{2}$

$\Rightarrow$ Path difference, $\text{x}=\frac{\lambda}{4}$

$\Rightarrow\text{y}=\frac{\text{xD}}{\text{d}}=\frac{\lambda\text{D}}{4\text{d}}$

  1. When intensity is $\frac{1}{4}\text{th}$ of the maximum $\frac{\text{I}}{\text{I}_\text{max}}=\frac{1}{4}$

$\Rightarrow\frac{4\text{a}^2\cos^2\Big(\frac{\phi}{2}\Big)}{4\text{a}^2}=\frac{1}{4}$

$\Rightarrow\cos^2\Big(\frac{\phi}{2}\Big)=\frac{1}{4}\Rightarrow\cos\Big(\frac{\phi}{2}\Big)=\frac{1}{2}$

$\Rightarrow\frac{\phi}{2}=\frac{\pi}{3}\Rightarrow\phi=\frac{2\pi}{3}$

$\Rightarrow$ Path difference, $\text{x}=\frac{\lambda}{3}$

$\Rightarrow\text{y}=\frac{\text{xD}}{\text{d}}=\frac{\lambda\text{D}}{3\text{d}}$

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