Question

Monochromatic light is incident on a pair of slits that are separated by 0.240 mm. The screen is 2.80 m away from the slits. (Assume the small-angle approximation is valid here.) (a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.62 cm, find the wavelength of the incident light.

Answer 1

Answer:

λ = 1.388 x 10⁻⁶ m = 1388 nm

Explanation:

We can use the Young's Double Slit experiment formula here to solve this numerical:

$\Delta x = \frac{\lambda L}{d} \\\\\lambda = \frac{\Delta x d}{L}$

where,

λ = wavelength = ?

Δx = distance between adjacent bright fringes = 1.62 cm = 0.0162 m

d = slit separation = 0.24 mm = 0.00024 m

L = distance from screen to slits = 2.8 m

Therefore,

$\lambda = \frac{(0.0162\ m)(0.00024\ m)}{2.8\ m}$

λ = 1.388 x 10⁻⁶ m = 1388 nm