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Chapter 34: Problem 38

An instructor uses light of wavelength \(633 \mathrm{nm}\) to create a diffraction pattern with a slit of width \(0.135 \mathrm{~mm} .\) How far away from the slit must the instructor place the screen in order for the full width of the central maximum to be \(5.00 \mathrm{~cm} ?\)

Short Answer

Expert verified
Answer: The distance between the slit and the screen should be approximately 5319.15 meters.

Step by step solution

01

1. Understand the problem and given values

We are given the wavelength of light (λ), the width of the slit (a), and the width of the central maximum (W). We need to find the distance (D) between the slit and the screen. Given values: λ = 633 nm a = 0.135 mm W = 5.00 cm First, let's convert these values into consistent units (meters). λ = 633 * 10^{-9} m a = 0.135 * 10^{-3} m W = 5.00 * 10^{-2} m
02

2. Derive the formula for the width of the central maximum

We can use the formula for single-slit diffraction patterns, which states that the angular width of the central maximum (θ) is given by: θ = 2 * arctan(λ / (2 * a))
03

3. Determine the relationship between θ, W, and D using the small-angle approximation

We can use the small-angle approximation, sin(θ) ≈ tan(θ) ≈ θ, to relate the angular width of the central maximum (θ), the width of the central maximum (W), and the distance between the slit and the screen (D): θ ≈ (W / 2) / D
04

4. Express the formula for the distance between the slit and the screen

We can re-write the small-angle approximation and diffraction pattern formulas to solve for the distance between the slit and the screen (D): D ≈ (W / 2) / θ = (W / 2) / [2 * arctan(λ / (2 * a))]
05

5. Calculate the distance D using the given values

Now we can plug the values for λ, a, and W into our formula and calculate the distance (D): D ≈ [(5.00 * 10^{-2} m) / 2] / [2 * arctan((633 * 10^{-9} m) / (2 * 0.135 * 10^{-3} m))] D ≈ (2.50 * 10^{-2} m) / [2 * arctan(2.34 * 10^{-6} m)] D ≈ (2.50 * 10^{-2} m) / 4.70 * 10^{-6} m D ≈ 5319.15 m
06

6. Conclusion

The instructor must place the screen approximately 5319.15 meters away from the slit in order for the full width of the central maximum to be 5.00 cm.

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Determine the minimum thickness of a soap film \((n=1.32)\) that would produce constructive interference when illuminated by light of wavelength of \(550 . \mathrm{nm} .\)

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