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3 years ago

11

A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance betw

een the first and second minima in the diffraction pattern is 4.75 mm. What is the width (in mm) of the slit?

1 answer:

mario62 [17]3 years ago

3 0

Answer:

$0.2$ mm

Explanation:

As we know that

$Y = \frac{m\lambda * D}{d}$

where

m represents the order of minimum

y represents the distance on the screen of the minimum from central axis

λ is the wavelength of the light

D is the distance between screen-to-slit

d represents the width of the slit

For first minima

$y_1 = \frac{1 * 633 * 10^{-9}*1.5}{d}$

For second minima

$y_2 = \frac{2 * 633 * 10^{-9}*1.5}{d}$

$Y_2 - Y_1 = 0.00475m\\\frac{633 * 10^{-9} * 1.5 }{d} = 0.00475\\d = 0.0002 m\\d = 0.2 mm$

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