3 years ago

8. 2 AI 6 HCI AICI: + H 에 싶

Physics

1 answer:

padilas [110]3 years ago

5 0

Answer:

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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

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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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