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

The angular position of the first-order, two-slit interference maxima is 1.26 degrees, and the angular position of the second-order maxima is 2.52 degrees. The intensity at the angular position of θ1 and θ2 can be calculated using the formula I = I0 cos^2(πy/λL) with the given values.

Step-by-step explanation:

To find the angular position of the first-order, two-slit interference maxima, we can use the formula θ = λ / d, where θ is the angular position, λ is the wavelength, and d is the slit separation. Plugging in the given values, we get θ1 = (580 nm) / (0.460 mm) = 1.26 degrees.

For the second-order maxima, we use the formula θ = 2λ / d. Plugging in the values, we get θ2 = 2(580 nm) / (0.460 mm) = 2.52 degrees.

The intensity at the angular position θ1 can be found using the formula I = I0 cos^2(πy/λL), where I0 is the intensity at the center, y is the distance from the center, λ is the wavelength, and L is the distance to the screen. Substituting the given values, we can calculate the intensity at θ1.

The intensity at the angular position θ2 can also be found using the same formula. Substituting the given values, we can calculate the intensity at θ2.