Question

An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.500 mm .

a. If the slits are very narrow, what would be the angular position of the second- order, two-slit interference maxima?
b. Let the slits have a width 0.300 mm. In terms of the intensity lo at the center of the central maximum, what is the intensity at the angular position in part "a"?

Answer 1

a

`\theta = 0.0022 rad`

b

`I = 0.000304 I_o`

Step-by-step explanation:

From the question we are told that

The wavelength of the light is
`\lambda = 550 \ nm = 550 *10^(-9) \ m`

The distance of the slit separation is
`d = 0.500 \ mm = 5.0 *10^(-4) \ m`

Generally the condition for two slit interference is

`dsin \theta = m \lambda`

Where m is the order which is given from the question as m = 2

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

substituting values

`\theta = 0.0022 rad`

Now on the second question

The distance of separation of the slit is

`d = 0.300 \ mm = 3.0 *10^(-4) \ m`

The intensity at the the angular position in part "a" is mathematically evaluated as

`I = I_o [(sin \beta)/(\beta) ]^2`

Where
`\beta` is mathematically evaluated as

`\beta = (\pi * d * sin(\theta ))/(\lambda )`

substituting values

`\beta = (3.142 * 3*10^(-4) * sin(0.0022 ))/(550 *10^(-9) )`

`\beta = 0.06581`

So the intensity is

`I = I_o [(sin (0.06581))/(0.06581) ]^2`

`I = 0.000304 I_o`