How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?

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Physics Gerda 3 years 2021-08-09T17:58:40+00:00 2021-08-09T17:58:40+00:00 1 Answers 64 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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Amity · Answer 1 · 2021-08-09T18:00:10+00:00

Answer:

$y = 0.0394 \ m$

Explanation:

From the question we are told that

The distance of the screen is $D = 2.20 \ m$

The distance of separation of the slit is $d = 0.0328 \ mm = 0.0328*10^{-3} \ m$

The wavelength of light is $\lambda = 588 \ nm = 588 *10^{-9} \ m$

Generally the condition for constructive interference is

$dsin\theta = n * \lambda$

=> $\theta = sin^{-1} [ \frac{ n * \lambda }{d } ]$

here n = 1 because we are considering the central diffraction peak

=> $\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ]$

=> $\theta = 1.0274 ^o$

Generally the width of central diffraction peak on a screen is mathematically evaluated as

$y = D tan (\theta )$

substituting values

$y = 2.20 * tan (1.0274)$

$y = 0.0394 \ m$