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timofeeve [1]
3 years ago
15

A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

each a point on a far-away screen such thatrays from the slit make an angle of 1.0° with the normal. Thedifference in phase for waves from the top and bottom of the slitis:
A) 0
B) 0.55 rad
C) 1.1 rad
D) 1.6 rad
E) 2.2 rad
Physics
1 answer:
kaheart [24]3 years ago
6 0

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

\Phi = \frac{2\pi \delta}{\lambda}

Where

\delta = Horizontal distance between two points

\lambda = Wavelength

From our values we have,

\lambda = 500nm = 5*10^{-6}m

\theta = 1\°

The horizontal distance between this two points would be given for

\delta = dsin\theta

Therefore using the equation we have

\Phi = \frac{2\pi \delta}{\lambda}

\Phi = \frac{2\pi(dsin\theta)}{\lambda}

\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}

\Phi= 1.096 rad \approx = 1.1 rad

Therefore the correct answer is C.

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