Question

Light from a 600 nm source goes through two slits 0.080 mm apart. What is the angular separation of the two first order maxima occurring on a screen 2.0 m from the slits

Answer 1

The angular separation is
`k = 0.8594^o`

Step-by-step explanation:

From the question we are told that

The wavelength of the light is
`\lambda = 600 \ nm = 600*10^(-9) \ m`

The distance of separation between the slit is
`d = 0.080 \ mm = 0.080 *10^(-3) \ m`

The distance from the screen is

Generally the condition for constructive interference is mathematically represented as

`d \ sin(\theta) = n \lambda`

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

here
`\theta` is the angular separation between the central maxima and one side of the first order maxima

given that we are considering the first order of maxima n = 1

=>
`\theta = sin ^(-1) [ (1 * 600*10^(-9) )/( 2.0 ) ]`

=>
`\theta = sin ^(-1) [ 0.0075 ]`

=>
`\theta = 0.4297^o`

So the angular separation of the two first order maxima is

`k = 2 * \theta`

`k = 2 * 0.4297`

`k = 0.8594^o`