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Chapter 34: Problem 21

What minimum path difference is needed to cause a phase shift by \(\pi / 4\) in light of wavelength \(700 . \mathrm{nm} \)

Short Answer

Expert verified
Answer: The minimum path difference required is approximately 87.5 nm.

Step by step solution

01

Understand the relationship between phase angle, wavelength, and path difference

The relationship between phase shift (\(\Delta \phi\)), wavelength (\(\lambda\)), and the path difference (\(\delta\)) is given by the equation: \(\Delta \phi = \cfrac{2 \pi \delta}{\lambda}\) Here, we are given the phase shift \(\Delta \phi = \cfrac{\pi}{4}\) and wavelength, \(\lambda=700 \ \mathrm{nm}\). We will now solve for the path difference \(\delta\).
02

Solve for path difference

We can isolate the path difference (\(\delta\)) by rewriting the equation in Step 1: \(\delta = \cfrac{\lambda \Delta \phi}{2 \pi}\) Next, plug in the given values of phase shift and wavelength: \(\delta = \cfrac{(700 \ \mathrm{nm})\left(\cfrac{\pi}{4}\right)}{2 \pi}\)
03

Calculate the path difference

Now we can calculate the minimum path difference for the given situation: \(\delta = \cfrac{(700 \ \mathrm{nm})\left(\cfrac{\pi}{4}\right)}{2 \pi} = \cfrac{700}{8} \ \mathrm{nm}\) \(\delta \approx 87.5 \ \mathrm{nm}\) Therefore, the minimum path difference required to cause a phase shift of \(\pi/4\) for light with a wavelength of \(700 \ \mathrm{nm}\) is approximately \(87.5 \ \mathrm{nm}\).

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Most popular questions from this chapter

Think of the pupil of your eye as a circular aperture \(5.00 \mathrm{~mm}\) in diameter. Assume you are viewing light of wavelength \(550 \mathrm{nm}\), to which your eyes are maximally sensitive. a) What is the minimum angular separation at which you can distinguish two stars? b) What is the maximum distance at which you can distinguish the two headlights of a car mounted \(1.50 \mathrm{~m}\) apart?

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