A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Exercise
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Given,
Distance of the screen from the slit, $\mathrm{D}=1 \mathrm{~m}$
Distance of the first minimum when the centre of screen, $x=2.5 \mathrm{~mm}=2.5 \times 10^{-3} \mathrm{~m}$
$\mathrm{n}=1$ is the first minimum.
Wavelength of light, $\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}$
$=5 \times 10^{-7} \mathrm{~m}$
Using formula, $\text{x}=\text{n}\frac{\lambda\text{D}}{\text{d}}$
$\text{we have},\ \text{d}=\text{n}\frac{\lambda\text{D}}{\text{x}}$
$\Rightarrow\ \text{d}=\frac{1\times5\times10^{-7}\times1}{2.5\times10^{-3}}$
$= 2 × 10^{-4}\ m$
$= 0.2\ mm$
is the required width of the slit.
Distance of the screen from the slit, $\mathrm{D}=1 \mathrm{~m}$
Distance of the first minimum when the centre of screen, $x=2.5 \mathrm{~mm}=2.5 \times 10^{-3} \mathrm{~m}$
$\mathrm{n}=1$ is the first minimum.
Wavelength of light, $\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}$
$=5 \times 10^{-7} \mathrm{~m}$
Using formula, $\text{x}=\text{n}\frac{\lambda\text{D}}{\text{d}}$
$\text{we have},\ \text{d}=\text{n}\frac{\lambda\text{D}}{\text{x}}$
$\Rightarrow\ \text{d}=\frac{1\times5\times10^{-7}\times1}{2.5\times10^{-3}}$
$= 2 × 10^{-4}\ m$
$= 0.2\ mm$
is the required width of the slit.
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