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QuestionA slit 248μm248 \mu m wide is illuminated by light of wavelength 591 nm . A diffraction pattern appears on a screen 4.5 m away. On the screen, what is the distance between the first and second diffraction minima on one side of the central maximum?
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Light of wavelength 574 nm passes through a slit, creating a diffraction pattern on a screen. The angular separation between the first diffraction minima on either side of the central maximum is 4.3∘4.3^{\circ}. Determine the width of the slit.
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Studdy Solution

STEP 1

1. The diffraction pattern is described by the single-slit diffraction formula.
2. The angle for the minima in a single-slit diffraction pattern is given by asin⁡θ=mλ a \sin \theta = m \lambda , where a a is the slit width, θ \theta is the angle of the minima, m m is the order of the minima, and λ \lambda is the wavelength of the light.
3. We are interested in the distance between the first and second diffraction minima on one side of the central maximum.

STEP 2

1. Calculate the angle for the first and second minima.
2. Determine the position of the minima on the screen.
3. Calculate the distance between the first and second minima.

STEP 3

For the first diffraction minimum (m=1 m = 1 ), use the formula:
asin⁡θ1=1⋅λ a \sin \theta_1 = 1 \cdot \lambda
Given: - a=248×10−6 m a = 248 \times 10^{-6} \, \text{m} - λ=591×10−9 m \lambda = 591 \times 10^{-9} \, \text{m}
Solve for θ1 \theta_1 :
sin⁡θ1=λa=591×10−9248×10−6 \sin \theta_1 = \frac{\lambda}{a} = \frac{591 \times 10^{-9}}{248 \times 10^{-6}}

STEP 4

Calculate θ1 \theta_1 :
sin⁡θ1=591×10−9248×10−6≈0.002383 \sin \theta_1 = \frac{591 \times 10^{-9}}{248 \times 10^{-6}} \approx 0.002383
θ1≈arcsin⁡(0.002383) \theta_1 \approx \arcsin(0.002383)

STEP 5

For the second diffraction minimum (m=2 m = 2 ), use the formula:
asin⁡θ2=2⋅λ a \sin \theta_2 = 2 \cdot \lambda
Solve for θ2 \theta_2 :
sin⁡θ2=2⋅591×10−9248×10−6 \sin \theta_2 = \frac{2 \cdot 591 \times 10^{-9}}{248 \times 10^{-6}}

STEP 6

Calculate θ2 \theta_2 :
sin⁡θ2=2⋅591×10−9248×10−6≈0.004766 \sin \theta_2 = \frac{2 \cdot 591 \times 10^{-9}}{248 \times 10^{-6}} \approx 0.004766
θ2≈arcsin⁡(0.004766) \theta_2 \approx \arcsin(0.004766)

STEP 7

Determine the position of the minima on the screen using the small angle approximation y≈Ltan⁡θ≈Lsin⁡θ y \approx L \tan \theta \approx L \sin \theta , where L=4.5 m L = 4.5 \, \text{m} .
For the first minimum:
y1=Lsin⁡θ1 y_1 = L \sin \theta_1
For the second minimum:
y2=Lsin⁡θ2 y_2 = L \sin \theta_2

STEP 8

Calculate y1 y_1 and y2 y_2 :
y1=4.5×0.002383≈0.010724 m y_1 = 4.5 \times 0.002383 \approx 0.010724 \, \text{m}
y2=4.5×0.004766≈0.021447 m y_2 = 4.5 \times 0.004766 \approx 0.021447 \, \text{m}

STEP 9

Calculate the distance between the first and second minima:
Δy=y2−y1 \Delta y = y_2 - y_1
Δy=0.021447−0.010724 \Delta y = 0.021447 - 0.010724

STEP 10

Calculate Δy \Delta y :
Δy≈0.010723 m \Delta y \approx 0.010723 \, \text{m}
The distance between the first and second diffraction minima on one side of the central maximum is approximately:
0.010723 m \boxed{0.010723 \, \text{m}}

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