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In a Young's double slit experiment, if the wavelength of light is increased by $50 \%$ and the distance between the slits is doubled then the percentage change in fringe width is
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The correct answer is:
$75$
Wavelength of light, $\lambda^{\prime}=50 \lambda$
Distance between the slit, $\mathrm{d}^{\prime}=2 \mathrm{~d}$ fringe width, $\beta^{\prime}=\frac{D \lambda^{\prime}}{d^{\prime}}$
$\begin{aligned}
& =\frac{D \times 0.50 \lambda}{2 d}=0.25 \beta \\
& \left|\frac{\beta^{\prime}-\beta}{\beta}\right| \times 100 \%=|0.25-1| \times 100 \% \\
& =75 \% .
\end{aligned}$
Distance between the slit, $\mathrm{d}^{\prime}=2 \mathrm{~d}$ fringe width, $\beta^{\prime}=\frac{D \lambda^{\prime}}{d^{\prime}}$
$\begin{aligned}
& =\frac{D \times 0.50 \lambda}{2 d}=0.25 \beta \\
& \left|\frac{\beta^{\prime}-\beta}{\beta}\right| \times 100 \%=|0.25-1| \times 100 \% \\
& =75 \% .
\end{aligned}$
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