Question

Light of wavelength 750 nm passes through a slit 1.0 μm wide and a single-slit diffraction pattern is formed vertically on a screen 34 cm away.

Determine the light intensity I 14 cm above the central maximum, expressed as a fraction of the central maximum's intensity I0.

Answer 1

Answer 2

To determine the light intensity at a certain point in a single-slit diffraction pattern, we can use the equation for the intensity of a single-slit diffraction pattern:

I = I0 * (sin(θ) / θ)^2 * (sin(N * π * d / λ) / (N * π * d / λ))^2

where:

• I is the intensity at a given point,

• I0 is the intensity at the central maximum,

• θ is the angle between the central maximum and the point,

• N is the number of bright fringes from the central maximum to the point,

• d is the width of the slit,

• λ is the wavelength of the light.

In this case, we are interested in the intensity 14 cm above the central maximum. Let's calculate the values needed to plug into the equation:

θ = arctan(y / L) where y is the distance above the central maximum (14 cm) and L is the distance between the slit and the screen (34 cm).

N = y * λ / (d * L) where y is the distance above the central maximum (14 cm), λ is the wavelength of the light (750 nm), d is the width of the slit (1.0 μm), and L is the distance between the slit and the screen (34 cm).

Now we can calculate the intensity at the given point using the provided values:

θ = arctan(0.14 m / 0.34 m) ≈ 0.389 N = (0.14 m * 750 nm) / (1.0 μm * 0.34 m) ≈ 3.456

I = I0 * (sin(θ) / θ)^2 * (sin(N * π * d / λ) / (N * π * d / λ))^2

Remember that the fraction of intensity is given by I / I0.

Please note that this calculation assumes the single-slit diffraction pattern is in the far-field (Fraunhofer diffraction) and the screen size is large compared to the pattern.

Substituting the values into the equation, we can determine the intensity at the given point as a fraction of the central maximum's intensity I0.