5. A single slit of width 0.05 mm is illuminated with monochromatic light of wavelength 500nm. Calculate the intensity of light at an angle of 20∘ relative to the central maximum on a distant screen.
Question
- A single slit of width 0.05 mm is illuminated with monochromatic light of wavelength 500nm. Calculate the intensity of light at an angle of 20∘ relative to the central maximum on a distant screen.
Solution
To solve this problem, we need to use the formula for the intensity of light in a single-slit diffraction pattern:
I(θ) = I0 * (sin(β)/β)²
where I0 is the maximum intensity, θ is the angle relative to the central maximum, and β is given by:
β = π * a * sin(θ) / λ
where a is the slit width, and λ is the wavelength of the light.
Step 1: Convert all quantities to the same units. Here, we'll use meters:
a = 0.05 mm = 0.05 * 10^-3 m λ = 500 nm = 500 * 10^-9 m θ = 20° = 20 * π/180 rad = 0.349 rad (since we need to convert degrees to radians)
Step 2: Calculate β:
β = π * a * sin(θ) / λ = π * 0.05 * 10^-3 * sin(0.349) / 500 * 10^-9 = 6.62
Step 3: Calculate the intensity:
I(θ) = I0 * (sin(β)/β)² = I0 * (sin(6.62)/6.62)² = 0.047 * I0
So, the intensity of light at an angle of 20° relative to the central maximum is approximately 4.7% of the maximum intensity.
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