Question

You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum

Answer 1

The width is
`Z = 0.0424 \ m`

Step-by-step explanation:

From the question we are told that

The width of the slit is
`d = 77.7 \mu m = 77.7 *10^(-6) \ m`

The wavelength of the light is
`\lambda = 721 \ nm`

The position of the screen is
`D = 2.83 \ m`

Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as

`\theta = sin ^(-1)[(m \lambda)/(d) ]`

Where m which is the order of the interference is 1

substituting values

`\theta = sin ^(-1)[(1 *721*10^(-9))/( 77.7*10^(-6)) ]`

`\theta = 0.5317 ^o`

Now the width of first minimum of the interference pattern is mathematically evaluated as

`Y = D sin \theta`

substituting values

`Y = 2.283 * sin (0.5317)`

`Y = 0.02 12 \ m`

Now the width of the pattern's central maximum is mathematically evaluated as

`Z = 2 * Y`

substituting values

`Z = 2 * 0.0212`

`Z = 0.0424 \ m`