mean: a double-slit interference experiment is set up in a laboratory using a source of blue monochromatic light of wavelength 475 nm.the separation of two slit is 0.40mm and the distance from the slit to the screen where the fringes are observed is 2.20m

Isn't there a standard formula for this? You have to convert separation and distance from the slits to an angle, but geepers, that is not onerous.

To calculate the positions of the interference fringes in a double-slit interference experiment, we can use the formula:

y = (λL) / d

where:
- y is the distance from the central maximum to a specific fringe on the screen
- λ is the wavelength of the light
- L is the distance from the slits to the screen
- d is the separation between the two slits

Given:
- λ (wavelength) = 475 nm = 475 × 10⁻⁹ m
- d (separation between slits) = 0.40 mm = 0.40 × 10⁻³ m
- L (distance from slits to screen) = 2.20 m

Substituting these values into the formula, we can determine the distance between the central maximum and the fringes on the screen:

y = [(475 × 10⁻⁹ m) × (2.20 m)] / (0.40 × 10⁻³ m)

Calculating:

y ≈ 2.61 × 10⁻³ m

Therefore, the distance between the central maximum and a specific fringe on the screen is approximately 2.61 × 10⁻³ meters.

To understand the behavior of light in a double-slit interference experiment, we need to calculate the fringe separation or the distance between two adjacent bright or dark fringes.

The formula for fringe separation in a double-slit interference experiment is given by:

d = λL / D

Where:
d is the fringe separation
λ is the wavelength of light
L is the distance from the slits to the screen
D is the separation of the two slits

In this case:
λ = 475 nm (converted to meters by dividing by 1 billion: 475 nm / 1,000,000,000 = 4.75 x 10^(-7) m)
L = 2.20 m
D = 0.40 mm (converted to meters by dividing by 1000: 0.40 mm / 1000 = 4 x 10^(-4) m)

Now, we can substitute these values into the formula to find the fringe separation:

d = (4.75 x 10^(-7) m)(2.20 m) / (4 x 10^(-4) m)
d ≈ 5.26 x 10^(-3) m

Therefore, the fringe separation is approximately 5.26 x 10^(-3) meters.