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Chapter 32: Problem 36

For a double-slit experiment with slit spacing \(0.25 \mathrm{mm}\) and wavelength \(600 \mathrm{nm},\) at what angular position is the path difference a quarter wavelength?

Short Answer

Expert verified
The angular position at which the path difference is a quarter wavelength is approximately \(0.0344\) degrees.

Step by step solution

01

Define Given Variables

Identify and write out the given variables. The slit spacing \(d\) is given as \(0.25 \mathrm{mm} = 2.5 \times 10^{-4} \mathrm{m}\). The wavelength \(\lambda\) is given as \(600 \mathrm{nm} = 6 \times10^{-7}\mathrm{m}\). The path difference is a quarter of the wavelength, hence \( \delta r = \lambda /4 = 1.5 \times 10^{-7} \mathrm{m}\).
02

Write Out the Equation for Path Difference

The path difference in a double-slit experiment is given by the equation \(\delta r = d \sin \theta\). This equation will be used to calculate the unknown \(\theta\).
03

Substitute Given Variables into Equation

Substitute the given variables into the path difference equation, which results in the following: \(1.5 \times 10^{-7} \mathrm{m} = 2.5 \times 10^{-4} \mathrm{m} \sin \theta\).
04

Solve for Unknown Angular Position

Rearrange the equation from Step 3 to solve for the angular position \(\theta\): \(\sin \theta = \frac{1.5 \times 10^{-7}}{2.5 \times 10^{-4}}\). Use your calculator to find the inverse sine of the right hand side to get \(\theta = \sin^{-1} \left(\frac{1.5 \times 10^{-7}}{2.5 \times 10^{-4}}\right) = 0.0006 \, radians\). Convert the result from radians to degrees by multiplying by \( \frac{180}{\pi}\) giving an approximate value of \(\theta = 0.0344 \, degrees\).

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Most popular questions from this chapter

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