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Chapter 25: Problem 31

Light incident on a pair of slits produces an interference pattern on a screen \(2.50 \mathrm{m}\) from the slits. If the slit separation is $0.0150 \mathrm{cm}$ and the distance between adjacent bright fringes in the pattern is \(0.760 \mathrm{cm},\) what is the wavelength of the light? [Hint: Is the small-angle approximation justified?]

Short Answer

Expert verified
Answer: The wavelength of the light is approximately 456 nm.

Step by step solution

01

List down given values

: Let's first write down the given values: - Distance between the screen and the slits (L) is \(2.50 \mathrm{m}\) - Slit separation (d) is \(0.0150 \mathrm{cm}\) - Distance between adjacent bright fringes (y) is \(0.760 \mathrm{cm}\) Now, let's convert the distances into the same unit (preferably meters). The slit separation, d = \(0.0150 \mathrm{cm} \times 10^{-2} = 1.50 \times 10^{-4} \mathrm{m}\) and the distance between adjacent bright fringes, y = \(0.760 \mathrm{cm} \times 10^{-2} = 7.60 \times 10^{-3} \mathrm{m}\).
02

Validate whether the small-angle approximation is justified

: The small-angle approximation is justified if the angle θ is very small, which means \(\tan(\theta) \approx \sin(\theta)\). This approximation is valid if \(y<<L\). Comparing the given values, we have \(7.60 \times 10^{-3} \mathrm{m} << 2.50 \mathrm{m}\). Since \(y\) is much smaller than \(L\), the small-angle approximation is justified.
03

Apply the double slit interference formula and solve for the wavelength

: The formula for double slit interference is: \(\frac{y}{L} \approx m\frac{\lambda}{d}\) Where: - \(m\) is the order of the bright fringe (an integer), considering two adjacent bright fringes, we can take \(m=1\), - \(\lambda\) is the wavelength of the light. Now, plug in the given values and solve for \(\lambda\): \(\lambda = \frac{y \times d}{L}\) \(\lambda = \frac{(7.60 \times 10^{-3} \mathrm{m}) \times (1.50 \times 10^{-4} \mathrm{m})}{2.50 \mathrm{m}}\) \(\lambda \approx 4.56 \times 10^{-7} \mathrm{m}\) Now, convert the wavelength back to nanometers (nm): \(\lambda \approx 4.56 \times 10^{-7} \mathrm{m} \times 10^9 = 456 \mathrm{nm}\) The wavelength of the light is around \(456 \mathrm{nm}\).

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Most popular questions from this chapter

Thin Films At a science museum, Marlow looks down into a display case and sees two pieces of very flat glass lying on top of each other with light and dark regions on the glass. The exhibit states that monochromatic light with a wavelength of \(550 \mathrm{nm}\) is incident on the glass plates and that the plates are sitting in air. The glass has an index of refraction of \(1.51 .\) (a) What is the minimum distance between the two glass plates for one of the dark regions? (b) What is the minimum distance between the two glass plates for one of the light regions? (c) What is the next largest distance between the plates for a dark region? [Hint: Do not worry about the thickness of the glass plates; the thin film is the air between the plates.]
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