Green light (λ = 518 nm) strikes a single slit at normal incidence. what width slit will produce a central maximum that is 3.00 cm wide on a screen 2.00 m from the slit? answer in μm .

6.9066 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

....1

The location of the bright fringe is determined by :

Where, L is the distance between the slit and the screen.

For small angle ,

So,  

Formula becomes:

Using 1, we get:

Thus, the distance between the central maximum is 3.00 cm

First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm

Since,

1 cm = 0.01 m

y = 0.0150 m

Given L = 2.00 m

λ = 518 nm

Since, 1 nm = 10⁻⁹ m

So,

λ = 518 × 10⁻⁹ m

Applying the formula as:

⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m

answer: option c.

explanation:

k2s for potassium sulphide