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andre [41]
3 years ago
8

A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the

central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width W2), and a wavelength of λ2 = 496 nm is used. The width of the central bright fringe on the screen is observed to be unchanged. Find W2.
Physics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

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Answer:

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An inexperienced researcher runs an experiment and sets his alpha level at .40 because he can't wait to get his firstsignificant
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Answer:

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Explanation:

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