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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

A spectrometer is used to analyze a light source. The screen-to-grating distance is \(50.0 \mathrm{cm}\) and the grating has 5000.0 slits/cm. Spectral lines are observed at the following angles: $12.98^{\circ}, 19.0^{\circ}, 26.7^{\circ}, 40.6^{\circ}, 42.4^{\circ}\( \)63.9^{\circ},\( and \)77.6^{\circ} .$ (a) How many different wavelengths are present in the spectrum of this light source? Find each of the wavelengths. (b) If a different grating with 2000.0 slits/cm were used, how many spectral lines would be seen on the screen on one side of the central maximum? Explain.
About how close to each other are two objects on the Moon that can just barely be resolved by the \(5.08-\mathrm{m}-\) (200-in.)-diameter Mount Palomar reflecting telescope? (Use a wavelength of \(520 \mathrm{nm}\).)
A thin soap film \((n=1.35)\) is suspended in air. The spectrum of light reflected from the film is missing two visible wavelengths of $500.0 \mathrm{nm}\( and \)600.0 \mathrm{nm},$ with no missing wavelengths between the two. (a) What is the thickness of the soap film? (b) Are there any other visible wavelengths missing from the reflected light? If so, what are they? (c) What wavelengths of light are strongest in the transmitted light?
You are given a slide with two slits cut into it and asked how far apart the slits are. You shine white light on the slide and notice the first-order color spectrum that is created on a screen \(3.40 \mathrm{m}\) away. On the screen, the red light with a wavelength of 700 nm is separated from the violet light with a wavelength of 400 nm by 7.00 mm. What is the separation of the two slits?
Sketch a sinusoidal wave with an amplitude of \(2 \mathrm{cm}\) and a wavelength of \(6 \mathrm{cm} .\) This wave represents the electric field portion of a visible EM wave traveling to the right with intensity \(I_{0}\). (a) Sketch an identical wave beneath the first. What is the amplitude (in centimeters) of the sum of these waves? (b) What is the intensity of the new wave? (c) Sketch two more coherent waves beneath the others, one of amplitude \(3 \mathrm{cm}\) and one of amplitude \(1 \mathrm{cm},\) so all four are in phase. What is the amplitude of the four waves added together? (d) What intensity results from adding the four waves?
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