Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.380 mm wide. The diffraction pattern is observed on a screen 3.55 m away. Define the width of a b…

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asked Jul 4, 2020 27.5k views

1 Answer

Arnekolja · Answer 1 · 2020-07-09T18:49:59+0000

Answer:

width of maximum intensity at the center of screen is 12 mm

Step-by-step explanation:

Position of minimum intensity on the screen is given as

$a sin\theta = N\lambda$

now we know that for first position of minimum intensity on the screen we have

N = 1

now we know

$a sin\theta = 1 \lambda$

a = 0.380 mm

$\lambda = 633 nm$

now we have

$(0.380 * 10^(-3))sin\theta = (633 * 10^(-9))$

$\theta = 0.095  degree = 1.66 * 10^(-3) rad$

now total angular width of central maximum is given as

$\phi = 2\theta = 2(1.66 * 10^(-3)) rad$

$\phi = 3.33 * 10^(-3) rad$

now linear width is given as

$w = L\phi$

w = (3.55)(3.33 * 10^(-3))

w = 0.012 m = 12 mm