Light with a wavelength of 410 nm (4.1 x 10-7 m) is incident upon a double slit with a separation of 0.6 mm (6 x 10-4 m). a screen is location 3.6 m from the double slit. (a) at what distance from the center of the screen will the first bright fringe beyond the center fringe appear?

2.4599×10⁻³ m

Explanation:

The relation between the distance of the slits and the angle through which it is passed is:

dsinθ=nλ  

where,

n can be any integer

d is the distance between the two slits  

λ is the wavelength of light.

Since for the first bright fringe, n=1

The experiment scenario can be drawn with the help of an image shown below.

Given:

d = 6×10⁻⁴ m

λ  = 4.1×10⁻⁷ m

6×10⁻⁴×sinθ=(1) 4.1×10⁻⁷

sinθ = 6.8333×10⁻⁴

Since sinθ is very small, so, sinθ = tanθ

tanθ = 6.8333×10⁻⁴

Distance between slit and screen = 3.6 m

So,

tanθ = (The distance if the first bright fringe)/(Distance between slit and screen)

The distance if the first bright fringe = 6.8333×10⁻⁴×3.6 m = 2.4599×10⁻³ m