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Chapter 32: Problem 38

A tube of glowing gas emits light at \(550 \mathrm{nm}\) and \(400 \mathrm{nm}\). In a double-slit apparatus, what's the lowest-order 550 -nm bright fringe that will fall on a 400 -nm dark fringe, and what are the fringes' corresponding orders?

Short Answer

Expert verified
The lowest-order 550-nm bright fringe that will fall on a 400-nm dark fringe is the 4th-order, and the corresponding orders of the fringes are 4 and 5.

Step by step solution

01

Understand the relationships of the fringes

In a double-slit experiment, the position of every bright fringe or constructive interference \(y_m\) can be given by the formula \(d \sin \theta = m \lambda\), where \(\lambda\) is the wavelength of light, \(d\) is the spacing between the slits, \(\theta\) is the angle that the light ray makes with respect to the central axis, and \(m\) is the order of the bright fringe. Note that \(m\) is always an integer because we can only have whole numbers of wavelengths. For a dark fringe or destructive interference, the condition is \(d \sin \theta = (m+\frac{1}{2}) \cdot \lambda\). Therefore, the problem can be broken down to finding the values of \(m\) that will satisfy both these relationships.
02

Setting up the related equation

In this exercise, we're required to find an order of a bright 550-nm fringe that will coincide with a dark 400-nm fringe. Thus we can set \(d \sin \theta_1 = m_1 \cdot 550 nm\) and \(d \sin \theta_2 = (m_2 + \frac{1}{2}) \cdot 400 nm\). Since the same dark fringe coincides with a bright fringe, \(\theta_1 = \theta_2\). This leads to \(m_1 \cdot 550 nm = (m_2 + \frac{1}{2}) \cdot 400 nm\).
03

Calculate the corresponding order

Solve the equation for the corresponding fringe orders. \(m_1\) and \(m_2\) should be the lowest possible positive integers that can satisfy the equation. By solving, we find that \(m_1 = 4\) and \(m_2 = 5\).

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Most popular questions from this chapter

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