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Using Young's double slit experiment, a monochromatic light of wavelength $5000 \mathrm{\AA}$ produces fringes of fringe width $0.5 \mathrm{~mm}$ . If another monochromatic light of wavelength $6000 \mathrm{\AA}$ is used and the separation between the slits is doubled, then the new fringe width will be :
Option: 1
$0.5\mathrm{mm}$

Option: 2
$1.0\mathrm{mm}$

Option: 3
$0.6\mathrm{mm}$

Option: 4
$0.3\mathrm{mm}$

Answers (1)

best_answer

Fringe width ( $\omega$ ) $\mathrm{=\frac{D\lambda }{d}}$

Fringe width ( $\omega$ ) $\mathrm{\alpha \ \ \frac{\lambda }{d}}$

$\mathrm{\frac{\omega^{\prime}}{\omega}=\frac{\lambda^{'} / d'}{\lambda / d}=\frac{\lambda^{\prime}}{\lambda} \times \frac{d}{d^{\prime}}}$ $\mathrm{=\frac{6000}{5000} \times \frac{d}{2 d}}$ $=\mathrm{\frac{6}{10}=0.6}$

$\begin{aligned} \mathrm{\omega ^{\prime} =\omega \times 0.6} \\ \end{aligned}$ $\begin{aligned} \mathrm{=0.5 \times 0.6=0.3 \mathrm{~mm}} \\ \end{aligned}$

Hence, the correct answer is Option (4).

Posted by

Divya Prakash Singh

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