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lapo4ka [179]
3 years ago
14

Two slits are illuminated by a 363 nm light. The angle between the zeroth-order bright band at the center of the screen and the

fourth-order bright band is 14.9 ◦ . If the screen is 170 cm from the double-slit, how far apart is this bright band from the central peak
Physics
1 answer:
Lynna [10]3 years ago
6 0

Answer:

y = 0.44 m

Explanation:

As we know that path difference on the screen is given as

\Delta x = \frac{yd}{L}

now for constructive interference we know that

path difference = integral multiple of wavelength

so we have

N\lambda = \frac{yd}{L}

now for 4th maximum on the screen we can say

y = \frac{N\lambda L}{d}

here N = 4

L = 170 cm = 1.70 m

\lambda = 363 nm

also we know

path difference on screen = dsin\theta

4\lambda = dsin14.9

d = \frac{4\lambda}{sin14.9}

now we have

y = \frac{4(\lambda)(1.70)}{\frac{4\lambda}{sin14.9}}

y = 0.44 m

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