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Chapter 3: Problem 24

Find the largest wavelength of light falling on double slits separated by \(1.20 \mu \mathrm{m}\) for which there is a first-order maximum. Is this in the visible part of the spectrum?

Short Answer

Expert verified
The largest wavelength that produces a 1st-order maxima for a double-slit setup with a slit separation of 1.20 µm is \(\lambda = 1.20 \times 10^{-6} \mathrm{m}\), which is equal to 1200 nm. This wavelength is not within the visible part of the spectrum, as the visible spectrum ranges from 380 nm to 750 nm.

Step by step solution

01

Set up the formula for double-slit interference patterns

We can use the formula for the maxima in a double-slit interference pattern: \[d \sin{\theta} = m\lambda\] Here, d: the distance between the slits (given as 1.20 micrometers) m: the order of the maxima (given as the 1st order, so m = 1) λ: the wavelength of the light θ: the angle between the central maxima and the m-th maxima Since we need to find the largest wavelength, we would need to maximize sinθ.
02

Maximize sinθ

The maximum value of sinθ is 1. Therefore, we can set sinθ = 1 in the formula: \[d \sin{\theta} = \lambda\] \[1.20 \times 10^{-6} \mathrm{m} \times 1 = \lambda\]
03

Calculate the largest wavelength

Now, we can find the largest wavelength by solving the equation: \[\lambda = 1.20 \times 10^{-6} \mathrm{m}\] This is the largest wavelength that will create a 1st-order maxima for the given double-slit setup.
04

Determine if the largest wavelength is in the visible spectrum

The visible spectrum of light has wavelengths ranging from 380 nm to 750 nm. Since 1.20 µm is equal to 1200 nm, it does not fall within the visible range. Thus, the largest wavelength that produces a 1st-order maxima for the given double-slit setup is not within the visible part of the spectrum.

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