Question:
A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ______.
Solution:
$\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
$\lambda=\frac{\beta \mathrm{d}}{\mathrm{D}}$
$\lambda=\frac{6 \times 10^{-3} \times 10^{-3}}{10}$
$\lambda=6 \times 10^{-7} \mathrm{~m}=600 \times 10^{-9} \mathrm{~m}$
$\lambda=600 \mathrm{~nm}$
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