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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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A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the
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Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, T_{b} = 75^{\circ}C

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h = 20\ W/m^{2}K

Now,

Perimeter of the fin, p = \pi D = 0.005\pi \ m

Cross-sectional area of the fin, A = \frac{\pi}{4}D^{2}

A = \frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}

To calculate the heat transfer rate:

Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})

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m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237

Now,

Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W

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