Question

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 .

Answer 1

6.9066 × 10⁻⁵ m

Step-by-step explanation:

For constructive interference, the expression is:

`d* sin\theta=m* \lambda`

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

d is the distance between the slits.

The formula can be written as:

`sin\theta=\frac {\lambda}{d}* m` ....1

The location of the bright fringe is determined by :

`y=L* tan\theta`

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

For small angle ,
`sin\theta=tan\theta`

So,

Formula becomes:

`y=L* sin\theta`

Using 1, we get:

`y=L* \frac {\lambda}{d}* m`

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:

`0.0150\ m=2.00\ m* \frac {518* 10^(-9)\ m}{d}* 1`

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