light of wavelength 729 nm and observe the resulting diffraction pattern on a screen that is situated 2.43 m from the slit. What is the width w, in centimeters, of the pattern’s central maximum?
1 point Problem 04: You illuminate a slit with a width of 73.5 µm with a light of wavelength 729 nm and observe the resulting diffraction pattern on a screen that is situated 2.43 m from the slit. What is the width w, in centimeters, of the pattern’s central maximum?
The Correct Answer and Explanation is:
To determine the width of the central maximum in a single-slit diffraction pattern, we use the formula derived from the diffraction condition:w=2L⋅tan(θ)≈2L⋅sin(θ)(for small angles)w = 2L \cdot \tan(\theta) \approx 2L \cdot \sin(\theta) \quad \text{(for small angles)}w=2L⋅tan(θ)≈2L⋅sin(θ)(for small angles)
For a single slit, the angle θ\thetaθ to the first minimum is given by the condition:a⋅sin(θ)=λa \cdot \sin(\theta) = \lambdaa⋅sin(θ)=λ
Where:
- aaa = slit width = 73.5 µm = 73.5×10−673.5 \times 10^{-6}73.5×10−6 m
- λ\lambdaλ = wavelength of light = 729 nm = 729×10−9729 \times 10^{-9}729×10−9 m
- LLL = distance from slit to screen = 2.43 m
- www = total width of central maximum = distance between first minima on both sides = 2y2y2y
Step 1: Calculate sin(θ)\sin(\theta)sin(θ):sin(θ)=λa=729×10−973.5×10−6=0.00991837\sin(\theta) = \frac{\lambda}{a} = \frac{729 \times 10^{-9}}{73.5 \times 10^{-6}} = 0.00991837sin(θ)=aλ=73.5×10−6729×10−9=0.00991837
Step 2: Use small-angle approximation (sin(θ)≈tan(θ)≈θ\sin(\theta) \approx \tan(\theta) \approx \thetasin(θ)≈tan(θ)≈θ)tan(θ)≈θ≈0.00991837 radians\tan(\theta) \approx \theta \approx 0.00991837 \ \text{radians}tan(θ)≈θ≈0.00991837 radians
Step 3: Compute half-width y=L⋅tan(θ)y = L \cdot \tan(\theta)y=L⋅tan(θ)y=2.43⋅0.00991837=0.0240936 my = 2.43 \cdot 0.00991837 = 0.0240936 \ \text{m}y=2.43⋅0.00991837=0.0240936 m
Step 4: Compute full width of central maximum:w=2y=2⋅0.0240936=0.0481872 m=4.82 cmw = 2y = 2 \cdot 0.0240936 = 0.0481872 \ \text{m} = 4.82 \ \text{cm}w=2y=2⋅0.0240936=0.0481872 m=4.82 cm
Final Answer:4.82 cm\boxed{4.82 \ \text{cm}}4.82 cm
Explanation:
In single-slit diffraction, the central maximum is the brightest and widest part of the diffraction pattern. It spans the distance between the first minima on either side of the center. The position of these minima depends on the wavelength of the light and the width of the slit. The relationship asinθ=λa \sin\theta = \lambdaasinθ=λ describes where the first minimum occurs. Since the angle is typically small, we can apply the small-angle approximation, replacing tanθ\tan\thetatanθ with sinθ\sin\thetasinθ, which simplifies calculations. The total width on the screen is then calculated using the geometry of the triangle formed by the slit, screen, and first minimum. After converting all units to SI, applying the formulas systematically yields the final width in meters, which is then converted to centimeters.
