Question

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0310.mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum (θ=0∘) is 3.00×10−5 W/m2 Part A What is the intensity at a point on the screen that corresponds to θ=1.20∘ ? Express your answer with the appropriate units.

Answer 1

The intensity at a point on the screen corresponding to θ = 1.20∘ is 5.60 ×
`10^-8 W/m^2`.

Explanation:

To determine the intensity at θ = 1.20∘, we can use the equation for intensity in a single-slit diffraction pattern:

`\[I(\theta) = I_0 \cdot \left((\sin(\alpha))/(\alpha)\right)^2\]`

where
`\(I(\theta)\)` is the intensity at an angle
`\(\theta\), \(I_0\)` is the intensity at the center, and
`\(\alpha = (\pi \cdot w \cdot \sin(\theta))/(\lambda)\)`, with w being the slit width,
`\(\lambda\)` the wavelength, and \
`(\theta\)` the angle.

Given
`\(I_0 = 3.00 * 10^(-5)\) W/m^2, \(\lambda = 592\) nm \(= 592 * 10^(-9)\) m, \(w = 0.0310 * 10^(-3)\) m, and \(\theta = 1.20^\circ\)`, we can calculate:

`\[\alpha = (\pi \cdot 0.0310 * 10^(-3) \cdot \sin(1.20^\circ))/(592 * 10^(-9))\]`

`\[\alpha \approx 0.01004\]`

Now, substitute
`\(\alpha\)` into the intensity equation:

`\[I(1.20^\circ) = 3.00 * 10^(-5) * \left((\sin(0.01004))/(0.01004)\right)^2\]`

`\[I(1.20^\circ) \approx 5.60 * 10^(-8)\] W/m^2`

Therefore, the intensity at a point on the screen corresponding to θ = 1.20∘ is
`\(5.60 * 10^(-8)\) W/m^2`. As the angle increases from the central maximum, the intensity decreases due to the spreading of the wavefronts caused by diffraction. In this case, at θ = 1.20∘, the intensity diminishes significantly from the central maximum.

Answer 2

To find the intensity at a point on the screen that corresponds to θ=1.20∘, we can use the formula for the intensity of a single-slit diffraction pattern. First, we need to find the phase difference due to the slit width, and then we can calculate the intensity at the desired angle.

Step-by-step explanation:

To find the intensity at a point on the screen that corresponds to θ=1.20∘, we can use the formula for the intensity of a single-slit diffraction pattern:

I(θ) = I(0) * (sin(β)/β)^2

Where I(θ) is the intensity at a specific angle, I(0) is the intensity at θ=0∘, and β is the phase difference due to the slit width.

First, we need to find β:

β = (π * d * sin(θ))/λ

Where d is the slit width, λ is the wavelength, and θ is the angle of interest. Plugging in the given values, we have:

β = (π * 0.0310 mm * sin(1.20∘))/(592 nm)

Next, we can find the intensity at θ=1.20∘:

I(θ) = 3.00×10−5 W/m2 * (sin(β)/β)^2

Calculating this value will give us the required intensity at θ=1.20∘.