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In Young's, double slit experiment, green light is incident on two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
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Using blue light instead of green light
The formula for fringe width is $W=\frac{\lambda D}{d}$
$\therefore \quad \mathrm{W} \propto \lambda, \mathrm{W} \propto \mathrm{D}$ and $\mathrm{W} \propto \frac{1}{\lambda}$
If the distance from the screen is increased, the width will increase.
If the distance between the slit is decreased, the width will increase.
As the wavelength will decrease the distance between the fringes will decrease.
$\lambda_{\text {nd }}>\lambda_{\text {erex }}>\lambda_{\text {max }}$
$\therefore \quad$ Blue light should be used.
$\therefore \quad \mathrm{W} \propto \lambda, \mathrm{W} \propto \mathrm{D}$ and $\mathrm{W} \propto \frac{1}{\lambda}$
If the distance from the screen is increased, the width will increase.
If the distance between the slit is decreased, the width will increase.
As the wavelength will decrease the distance between the fringes will decrease.
$\lambda_{\text {nd }}>\lambda_{\text {erex }}>\lambda_{\text {max }}$
$\therefore \quad$ Blue light should be used.
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