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Slav-nsk [51]
4 years ago
8

Light with wavelength 633 nm is incident on a 3.00-μm-wide slit.Part AFind the angular width of the central peak in the diffract

ion pattern, taken as the angular separation between the first minima.

Physics
1 answer:
Semenov [28]4 years ago
6 0

Answer:

The angular width of the central peak is 24.2 degrees

Explanation:

The equation that describes the single-slit diffraction phenomenon is:

a\sin\theta=m\lambda (1)

with a the width of the slit, \theta the angular position of the minimum regarding the center of the screen where light is projected, m the order of the minimum and \lambda the wavelength. Solving (1) for \theta with m=1 that is the first minimum:

\theta=\arcsin(\frac{m\lambda}{a})=\arcsin(\frac{(1)(633\times10^{-9})}{3.0\times10^{-6}})\approx12.1 deg

See the figure below that the central peak is symmetric regarding the center of the screen, which implies that the angular width of the central peak is 2\theta =2(12.1 deg) = 24.2 deg

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Let's call the constant acceleration a.
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From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.

Hence P(t)=a*t^2/2+10t

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