Light of wavelength 580 nm is incident normally on a slit of width ' a '. The distance between slit and screen is 2.5 m and the distance of the second order maximum from the centre of the sc

Question

Light of wavelength 580 nm is incident normally on a slit of width ' a '. The distance between slit and screen is 2.5 m and the distance of the second order maximum from the centre of the screen is 14.5 mm in a diffraction pattern. The value of ' a ' is

TestHub · Accepted Answer

For a single slit diffraction pattern, the position of the $n$ -th secondary maximum on the screen is given by:

$y = \frac{( 2 n + 1 ) λ D}{2 a}$

For the second order maximum, $n = 2$ . Substituting this into the formula, we get:

$y = \frac{5 λ D}{2 a}$

Rearranging for the slit width $a$ :

$a = \frac{5 λ D}{2 y}$

Substituting the given values $λ = 580 \times 1 0^{- 9} m$ , $D = 2.5 m$ , and $y = 14.5 \times 1 0^{- 3} m$ :

$a = \frac{5 \times 580 \times 1 0 ^{- 9} \times 2.5}{2 \times 14.5 \times 1 0 ^{- 3}}$

$a = \frac{7250 \times 1 0 ^{- 9}}{29 \times 1 0 ^{- 3}}$

$a = 250 \times 1 0^{- 6} m$

$a = 0.25 \times 1 0^{- 3} m$

Answer: $0.25 \times 1 0^{- 3} m$

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