Question

In a Young's interference experiment, the two slits are separated by 0.135 mm, and the incident light includes two wavelengths: 1 = 540 nm (green) and 2 = 450 nm (blue). The overlapping interference patterns are observed on a screen 1.23 m from the slits. What is the distance between the central maximum and the first-order maximum for the green light?

Option 1: 0.036 m
Option 2: 0.045 m
Option 3: 0.027 m
Option 4: 0.054 m

Answer 1

The distance between the central maximum and the first-order maximum for the green light in Young's interference experiment is 0.045 m.

The answer is option ⇒ 2

Step-by-step explanation:

In Young's interference experiment, the distance between the central maximum and the first-order maximum for the green light can be calculated using the formula:

d = λD/d

Where d is the distance between the slits, λ is the wavelength of light, and D is the distance between the slits and the screen.

Substituting the given values, we have:

d = (540nm)(1.23m)/(0.135mm)

d = 0.045m

Therefore, the distance between the central maximum and the first-order maximum for the green light is 0.045m.

The answer is option ⇒ 2: 0.045 m