A Young's double slit apparatus has slits separated by 0.28mm and a screen 48cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light ($\lambda=700\text{nm}$ in vacuum). Find the fringe-width of the pattern formed on the screen.
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Given that, $d = 0.28mm = 0.28 × 10^{-3}m, D = 48cm = 0.48m, \lambda_\text{a}=700\text{nm}$ in vacuum
Let, $\lambda_\text{w}$ = wavelength of red light in water
Since, the fringe width of the pattern is given by,
$\beta=\frac{\lambda_\text{w}\text{D}}{\text{d}}$
$=\frac{525\times10^{-9}\times0.48}{0.28\times10^{-3}}=9\times10^{-4}\text{m}=0.90\text{mm}.$
Let, $\lambda_\text{w}$ = wavelength of red light in water
Since, the fringe width of the pattern is given by,
$\beta=\frac{\lambda_\text{w}\text{D}}{\text{d}}$
$=\frac{525\times10^{-9}\times0.48}{0.28\times10^{-3}}=9\times10^{-4}\text{m}=0.90\text{mm}.$
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