How wide is the central diffraction peak on a screen 2.30 m behind a 0.0368 mm wide slit illuminated by 588 nm light?
The Correct Answer and Explanation is:
To find the width of the central diffraction peak, we use the single-slit diffraction formula. The width of the central maximum is defined as the distance between the first minima on both sides of the central peak.
Formula:
θ=sin−1(λa)\theta = \sin^{-1}\left(\frac{\lambda}{a}\right)θ=sin−1(aλ)y=Ltan(θ)y = L \tan(\theta)y=Ltan(θ)
Since the angle is small, tan(θ)≈sin(θ)\tan(\theta) \approx \sin(\theta)tan(θ)≈sin(θ), so:y=Lλay = L \frac{\lambda}{a}y=Laλ
Then the full width of the central peak is:Width=2y=2Lλa\text{Width} = 2y = 2L \frac{\lambda}{a}Width=2y=2Laλ
Given:
- L=2.30 mL = 2.30 \, \text{m}L=2.30m (distance to screen)
- a=0.0368 mm=3.68×10−5 ma = 0.0368 \, \text{mm} = 3.68 \times 10^{-5} \, \text{m}a=0.0368mm=3.68×10−5m (slit width)
- λ=588 nm=5.88×10−7 m\lambda = 588 \, \text{nm} = 5.88 \times 10^{-7} \, \text{m}λ=588nm=5.88×10−7m (wavelength of light)
Plug into formula:
Width=2×2.30×5.88×10−73.68×10−5\text{Width} = 2 \times 2.30 \times \frac{5.88 \times 10^{-7}}{3.68 \times 10^{-5}}Width=2×2.30×3.68×10−55.88×10−7=4.60×5.88×10−73.68×10−5= 4.60 \times \frac{5.88 \times 10^{-7}}{3.68 \times 10^{-5}}=4.60×3.68×10−55.88×10−7=4.60×0.01598= 4.60 \times 0.01598=4.60×0.01598≈0.0735 m=7.35 cm\approx 0.0735 \, \text{m} = 7.35 \, \text{cm}≈0.0735m=7.35cm
Final Answer:
7.35 cm\boxed{7.35 \, \text{cm}}7.35cm
Explanation:
When light passes through a narrow slit, it diffracts and spreads out, forming a pattern of bright and dark bands on a screen. The central bright fringe, also called the central maximum, is the widest and brightest. The first dark fringes (minima) on each side mark the end of the central peak.
The angle to the first minimum in single-slit diffraction occurs where:asin(θ)=λa \sin(\theta) = \lambdaasin(θ)=λ
For small angles, sin(θ)≈tan(θ)≈y/L\sin(\theta) \approx \tan(\theta) \approx y / Lsin(θ)≈tan(θ)≈y/L, so the distance from the center to the first minimum is:y=λLay = \frac{\lambda L}{a}y=aλL
To find the total width of the central peak, we double this value since the pattern is symmetrical. Using the given slit width, wavelength, and screen distance, we find the central peak to be 7.35 cm wide on the screen.
