Question:

A beam of light of wavelength 600 nm passes through a narrow slit of width 0.1 mm. The resulting diffraction pattern is observed on a screen 2 meters away from the slit. Calculate the distance between the first and second minima in the diffraction pattern.

Answer:

Given, Wavelength of light, λ = 600 nm = 600 × 10^-9 m Width of the slit, a = 0.1 mm = 0.1 × 10^-3 m Distance to the screen, L = 2 m

The distance between the minima can be calculated using the formula for the angular position of the nth minima:

sin(θn) = nλ / a

Where, n = order of the minima λ = wavelength of the light a = width of the slit

Finding the angular position of the first minima: sin(θ1) = (1*600 × 10^-9 m) / (0.1 × 10^-3 m) sin(θ1) ≈ 0.006 θ1 ≈ sin^(-1)(0.006) θ1 ≈ 0.34°

Finding the angular position of the second minima: sin(θ2) = (2*600 × 10^-9 m) / (0.1 × 10^-3 m) sin(θ2) ≈ 0.012 θ2 ≈ sin^(-1)(0.012) θ2 ≈ 0.69°

Now, to find the distance between the first and second minima, we can use the small angle approximation:

d = L tan(θ2) - L tan(θ1) d ≈ 2 tan(0.69°) - 2 tan(0.34°) d ≈ 0.024 m

Therefore, the distance between the first and second minima in the diffraction pattern is approximately 0.024 meters.