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Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is \(\lambda_{\mathrm{A}}=645 \mathrm{nm} .\) The other wavelength is \(\lambda_{\mathrm{B}}\) and is unknown. On a viewing screen, the light with wavelengthV \(\lambda_{\mathrm{A}}=645 \mathrm{nm}\) produces its third-order bright fringe at the same place where the light with wavelength \(\lambda_{\mathrm{B}}\) produces its fourth dark fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown wavelength?

Short Answer

Expert verified
The unknown wavelength \( \lambda_B \) is 430 nm.

Step by step solution

01

Understand the conditions for bright and dark fringes

For bright fringes in a double-slit experiment, the condition is \( d \sin \theta = m \lambda \), where \( m \) is the order of the fringe (an integer), \( \lambda \) is the wavelength, \( d \) is the distance between the slits, and \( \theta \) is the angle of the fringe relative to the central maximum.For dark fringes, the condition is \( d \sin \theta = (m + 0.5) \lambda \) for minima. In our case, for the fourth dark fringe, \( m = 4 \).
02

Set up the equation for bright fringes

For the third-order bright fringe of the known wavelength \( \lambda_A = 645 \text{ nm} \), we use the equation:\[ d \sin \theta = 3 \times 645 \text{ nm} \]
03

Set up the equation for dark fringes

For the fourth dark fringe of the unknown wavelength \( \lambda_B \), we use the equation:\[ d \sin \theta = (4 + 0.5) \lambda_B = 4.5 \lambda_B \]
04

Equate the two equations for the same position

Since both fringes are at the same position on the screen,\[ 3 \times 645 \text{ nm} = 4.5 \lambda_B \]
05

Solve for the unknown wavelength \(\lambda_B\)

Rearrange the equation to solve for \( \lambda_B \):\[ \lambda_B = \frac{3 \times 645 \text{ nm}}{4.5} \]Calculate this to find \( \lambda_B \).
06

Calculate the result

Perform the calculation:\[ \lambda_B = \frac{3 \times 645}{4.5} = 430 \text{ nm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Interference Fringes
In a double-slit experiment, light waves pass through two slits and then overlap, creating a pattern of bright and dark spots. These spots are known as interference fringes. Bright fringes occur where the light waves add up or constructively interfere. Dark fringes appear where the waves cancel out or destructively interfere.
The path difference for bright fringes is a multiple of the light's wavelength, represented by the equation \(d \sin \theta = m \lambda\). Here, \(\lambda\) is the wavelength, \(d\) is the slit separation, \(\theta\) is the angle to the fringe, and \(m\) is a positive integer, or the order number of the fringe. On the other hand, dark fringes are found at positions where \(d \sin \theta = (m + 0.5) \lambda\).
Understanding these fringe positions is key to predicting where bright and dark bands show up, helping scientists measure wave properties like wavelength.
Deriving the Wavelength Equation
In optical physics, understanding how to derive the wavelength equation from fringe conditions is pivotal. This relationship allows us to calculate the unknown wavelength when given conditions about interference fringes.
Consider the situation where light with a known wavelength and an unknown wavelength create overlapping fringes. We use the bright fringe condition \(d \sin \theta = m \lambda\) and the dark fringe condition \(d \sin \theta = (m + 0.5) \lambda\). By equating these expressions for a shared position on the viewing screen, one can find the unknown wavelength.
For instance, if the third bright fringe of a wave occurs at the same spot as the fourth dark fringe of another wave, we set their equations equal: \(3 \times 645 \text{ nm} = 4.5 \lambda_B\). Solving for \(\lambda_B\) gives us \(\lambda_B = \frac{3 \times 645 \text{ nm}}{4.5}\), which can be simplified to find the exact wavelength.
Exploring Optical Physics
Optical physics is the study of light and its interactions with matter. It's a fascinating field that explores how light behaves and how it can be manipulated. A critical area of investigation in optical physics is interference, which plays a significant role in experiments like the double-slit experiment.
Interference occurs when different light waves overlap and interact. This principle helps us understand phenomena such as diffraction and the behavior of lenses. Optical physics uses mathematical models and experimental techniques to explain these interactions and is essential in fields like photography, medical imaging, and telecommunications.
Studying optical physics not only reveals light's properties but also enables technological advancements that enhance our daily lives. The double-slit experiment exemplifies how subtle manipulations of light can lead to significant discoveries about wave behavior and fundamental physics.

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Most popular questions from this chapter

Two diffraction gratings, A and B, are located at the same distance from the observation screens. Light with the same wavelength \(\lambda\) is used for each. The separation between adjacent principal maxima for grating A is \(2.7 \mathrm{cm},\) and for grating \(\mathrm{B}\) it is \(3.2 \mathrm{cm} .\) Grating \(\mathrm{A}\) has 2000 lines per meter. How many lines per meter does grating \(\mathrm{B}\) have? (Hint: The diffraction angles are small enough that the approximation \(\sin \theta \approx \tan \theta \operatorname{can}\) be used. \()\)

A diffraction pattern forms when light passes through a singleslit. The wavelength of the light is 675 nm. Determine the angle that locates the first dark fringe when the width of the slit is (a) \(1.8 \times 10^{-4} \mathrm{m}\) and (b) \(1.8 \times 10^{-6} \mathrm{m}\)

In a Young's double-slit experiment, two rays of monochromatic light emerge from the slits and meet at a point on a distant screen, as in Figure \(27.6 a .\) The point on the screen where these two rays meet is the eighth-order bright fringe. The difference in the distances that the two rays travel is \(4.57 \times 10^{-6} \mathrm{m} .\) What is the wavelength (in \(\mathrm{nm}\) ) of the monochromatic light?

The wavelength of the laser beam used in a compact disc player is \(780 \mathrm{nm}\). Suppose that a diffraction grating produces first-order tracking beams that are \(1.2 \mathrm{mm}\) apart at a distance of \(3.0 \mathrm{mm}\) from the grating. Estimate the spacing between the slits of the grating.

In a Young's double-slit experiment, the wavelength of the light used is \(520 \mathrm{nm}\) (in vacuum), and the separation between the slits is \(1.4 \times 10^{-6} \mathrm{m}\). Determine the angle that locates (a) the dark fringe for which \(m=0,\) (b) the bright fringe for which \(m=1,\) (c) the dark fringe for which \(m=1,\) and (d) the bright fringe for which \(m=2\).

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