Question

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 3rd 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.

Answer 1

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

$d sin\theta = 3 \lambda$

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

$\theta = 0.311 degree$