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Chapter 3: Problem 58

Suppose that the highest order fringe that can be observed is the eighth in a double-slit experiment where 550-nm wavelength light is used. What is the minimum separation of the slits?

Short Answer

Expert verified
The minimum separation of the slits in the double-slit experiment is approximately \(2.81 × 10^(-6) \) meters or 2.81 µm.

Step by step solution

01

Write the formula for angular separation in a double-slit experiment

The formula for angular separation in a double-slit experiment is given by: \(θ = mλ / D\) Where, \(θ\) is the angular separation, \(m\) is the order of the fringe (in this case, the highest order is 8), \(λ\) is the wavelength of the light (550 nm), and \(D\) is the distance between the slits.
02

Make the formula to find the slit separation

Since we're looking for the minimum slit separation, the fringes must fall on the very edge of the detectable area. So, the angular separation should be 90 degrees. We can use this information and rewrite the formula to solve for the slit separation: \(D = mλ / θ\)
03

Convert the wavelength to meters

The wavelength is given in nanometers (nm), so we should convert it to meters (m) to work with standard units: 550 nm = 550 × 10^(-9) m
04

Convert the angle to radians

Since the angle should be in radians, we convert 90 degrees to radians: \(θ = 90 × π / 180 = π / 2\) radians
05

Plug in the values and calculate the minimum slit separation

Now, we can plug in the values and find the minimum slit separation: \(D = (8 × 550 × 10^(-9)) / (π / 2)\) m \(D ≈ 2.81 × 10^(-6) \) m So, the minimum separation of the slits in the double-slit experiment is approximately \(2.81 × 10^(-6) \) meters or 2.81 µm.

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Most popular questions from this chapter

Fifty-one narrow slits are equally spaced and separated by \(0.10 \mathrm{mm}\). The slits are illuminated by blue light of wavelength \(400 \mathrm{nm} .\) What is angular position of the twenty-fifth secondary maximum? What is its peak intensity in comparison with that of the primary maximum?

What effect does increasing the wedge angle have on the spacing of interference fringes? If the wedge angle is too large, fringes are not observed. Why?

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