Question

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

Answer 1

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 = d$sin\theta$

$4\lambda = dsin14.9$

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

now we have

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

$y = 0.44 m$