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In Young's double slit experimental setup, if the wavelength alone is doubled, the bandwidth $\beta$ becomes
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Verified Answer
The correct answer is:
$2 \beta$
Fringe width, $\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
When $\lambda$ alone is doubled then fringe width becomes
$$
\beta^{\prime}=\frac{(2 \lambda) \mathrm{D}}{\mathrm{d}}=2 \beta
$$
When $\lambda$ alone is doubled then fringe width becomes
$$
\beta^{\prime}=\frac{(2 \lambda) \mathrm{D}}{\mathrm{d}}=2 \beta
$$
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