10.19 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.

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Payal Gupta | Contributor-Level 10

9 months ago

10.19 Wavelength of the light beam, $?$ = 500 nm = 500 $* 10^{- 9}$ m

Distance of the screen from the slit, D = 1 m $n$

Distance of the first minimum from the centre of the screen, x = 2.5 mm = 2.5 $* 10^{- 3}$ m

Let the width of the slit be = d

From the equation

$n ? = x \frac{d}{D}$ we get d = $\frac{n ? D}{x}$

$d = \frac{1 * 500 * 10^{- 9} * 1}{2.5 * 10^{- 3}}$ = 2 $* 10^{- 4}$ m = 0.2 mm

Hence, the width of the slot is 0.2 m

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Physics Ncert Solutions Class 12th 2023

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