Question

An interference pattern is produced by light with a wavelength 590 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.580 mm .
Required:

a. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?

b. What would be the angular position of the second-order, two-slit, interference maxima in this case?

Answer 1

Answer:

a. 0.058°

b.  0.117°

Explanation:

a. The angular position of the first-order is:

$d*sin(\theta) = m\lambda$

$\theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ}$

Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.

b. The angular position of the second-order is:

$\theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ}$

Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.

I hope it helps you!