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alekssr [168]
3 years ago
9

Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3

rd order bright band on the screen is separated from the central maximum by 0.76 cm . A)Determine the frequency of the light.
B)Determine the slit separation.
C)Determine the angle of the third-order bright band.
Physics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

\lambda = 630 nm

now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

f = \frac{3 \times 10^8}{630 \times 10^{-9}}

f = 4.76 \times 10^{14} Hz

Part b)

Position of Nth maximum intensity on the screen is given as

y_n = \frac{n\lambda L}{d}

so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

n = 3

L = 1.4 m

0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

d = 3.48 \times 10^{-4} m

Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})

\theta = 0.311 degree

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