Question

How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?

Answer 1

`y = 0.0394 \ m`

Step-by-step explanation:

From the question we are told that

The distance of the screen is
`D = 2.20 \ m`

The distance of separation of the slit is
`d = 0.0328 \ mm = 0.0328*10^(-3) \ m`

The wavelength of light is
`\lambda = 588 \ nm = 588 *10^(-9) \ m`

Generally the condition for constructive interference is

`dsin\theta = n * \lambda`

=>
`\theta = sin^(-1) [ ( n * \lambda )/(d ) ]`

here n = 1 because we are considering the central diffraction peak

=>
`\theta = sin^(-1) [ ( 1 * 588*10^(-9) )/(0.0328*10^(-3) ) ]`

=>
`\theta = 1.0274 ^o`

Generally the width of central diffraction peak on a screen is mathematically evaluated as

`y = D tan (\theta )`

substituting values

`y = 2.20 * tan (1.0274)`

`y = 0.0394 \ m`