In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1º. What is the spacing between the two slits?
Exercise
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Wavelength of light, $\lambda=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}$
Angular fringe width, $\text{B}_\theta=\frac{\lambda}{\text{d}}$
$\Rightarrow\ \ \text{d}=\frac{\lambda}{\beta}_\theta$
Where, d is the distance between the slits.
Now,
$\beta_\theta=0.1^\circ=\frac{0.1\times\pi}{180}\ \text{radian}$
$=\frac{0.1\times3.14}{180}\ \text{radian}$
$\therefore\ \text{d}=\frac{600\times10^{-9}\times180}{0.1\times3.14}\ \text{m}=3.44\times10^{-4}\text{m}.$
Is the required spacing between the two slits.
Angular fringe width, $\text{B}_\theta=\frac{\lambda}{\text{d}}$
$\Rightarrow\ \ \text{d}=\frac{\lambda}{\beta}_\theta$
Where, d is the distance between the slits.
Now,
$\beta_\theta=0.1^\circ=\frac{0.1\times\pi}{180}\ \text{radian}$
$=\frac{0.1\times3.14}{180}\ \text{radian}$
$\therefore\ \text{d}=\frac{600\times10^{-9}\times180}{0.1\times3.14}\ \text{m}=3.44\times10^{-4}\text{m}.$
Is the required spacing between the two slits.
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