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Chapter 25: Problem 34

Light of wavelength 589 nm incident on a pair of slits produces an interference pattern on a distant screen in which the separation between adjacent bright fringes at the center of the pattern is \(0.530 \mathrm{cm} .\) A second light source, when incident on the same pair of slits, produces an interference pattern on the same screen with a separation of $0.640 \mathrm{cm}$ between adjacent bright fringes at the center of the pattern. What is the wavelength of the second source? [Hint: Is the small-angle approximation justified?]

Short Answer

Expert verified
In a double-slit interference experiment, the separation between adjacent bright fringes is determined by the difference in the position between consecutive orders and is given by the formula \(\Delta x = \dfrac{\lambda D}{d}\), where \(\lambda\) is the wavelength of the light, \(D\) is the distance from the screen to the slits, and \(d\) is the distance between the slits. Using the given separations between adjacent bright fringes for the first and second light sources, we can find the ratio of their wavelengths and solve for the unknown wavelength of the second light source. The wavelength of the second light source is found to be 711 nm.

Step by step solution

01

Understand the double-slit interference formula

The general formula for the position of bright fringes in a double-slit interference experiment is given by: \(x = \dfrac{m\lambda D}{d}\) where \(x\) is the position of the bright fringe on the screen, \(m\) is the order of the fringe (an integer), \(\lambda\) is the wavelength of the light, \(D\) is the distance from the screen to the slits, and \(d\) is the distance between the slits.
02

Find the relationship between the separations of adjacent bright fringes and the formula

The separation between adjacent bright fringes is determined by the difference in the position between consecutive orders, i.e., \((m+1)-m\). Using the formula from Step 1, the separation is given by: \(\Delta x = x_{m+1} - x_m = \dfrac{(m+1)\lambda D}{d} - \dfrac{m\lambda D}{d} = \dfrac{\lambda D}{d}\)
03

Use the given data to find the ratio of the wavelengths

We are given the separation between adjacent bright fringes for the first and second light sources. We can use the relationship we found in Step 2 to find the ratio of their wavelengths: \(\dfrac{\Delta x_1}{\Delta x_2} = \dfrac{\lambda_1}{\lambda_2}\) Plugging in the given values: \(\dfrac{0.530}{0.640} = \dfrac{589}{\lambda_2}\)
04

Solve for the unknown wavelength

From the equation in Step 3, we can solve for the wavelength of the second light source (\(\lambda_2\)): \(\lambda_2 = \dfrac{589 \times 0.640}{0.530} = 711 \mathrm{nm}\) The wavelength of the second light source is 711 nm.

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Most popular questions from this chapter

A soap film has an index of refraction \(n=1.50 .\) The film is viewed in transmitted light. (a) At a spot where the film thickness is $910.0 \mathrm{nm},$ which wavelengths are weakest in the transmitted light? (b) Which wavelengths are strongest in transmitted light?
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?]
Light from a red laser passes through a single slit to form a diffraction pattern on a distant screen. If the width of the slit is increased by a factor of two, what happens to the width of the central maximum on the screen?
If you shine a laser (wavelength \(0.60 \mu \mathrm{m}\) ) with a small aperture at the Moon, diffraction makes the beam spread out and the spot on the Moon is large. Making the aperture smaller only makes the spot on the Moon larger. On the other hand, shining a wide searchlight at the Moon can't make a tiny spot- the spot on the Moon is at least as wide as the searchlight. What is the radius of the smallest possible spot you can make on the Moon by shining a light from Earth? Assume the light is perfectly parallel before passing through a circular aperture.
In a double-slit experiment, what is the linear distance on the screen between adjacent maxima if the wavelength is \(546 \mathrm{nm}\), the slit separation is \(0.100 \mathrm{mm},\) and the slit-screen separation is \(20.0 \mathrm{cm} ?\)
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