Question

An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.460 mm .1. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?2. What would be the angular position of the second-order, two-slit, interference maxima in this case?3. Let the slits have a width 0.310 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of θ1?4. What is the intensity at the angular position of θ2?

Answer 1

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

Angular position of first order interference maxima

= λ / d radian

= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x  0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

I = I₀ cos²δ/2 ( δ is phase difference of two lights.

For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

=  I₀ cos².126x 10⁻²