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Chapter 34: Problem 40

Plane light waves are incident on a single slit of width \(2.00 \mathrm{~cm} .\) The second dark fringe is observed at \(43.0^{\circ}\) from the central axis. What is the wavelength of the light?

Short Answer

Expert verified
Answer: The wavelength of the light is approximately \(7.18 \times 10^{-7} \mathrm{~m}\).

Step by step solution

01

Determine the angle of the second dark fringe

First, we need to convert the given angle, \(43.0^{\circ}\), to radians, since we will be using the radians in subsequent calculations. To convert degrees to radians, we use the formula: $$ 1^{\circ} = \frac{\pi}{180} \: \text{radians} $$ Let \(\theta\) be the angle of the second dark fringe in radians. Then we have: $$ \theta = 43.0^{\circ} \times \frac{\pi}{180} = \frac{43\pi}{180} \: \text{radians} $$
02

Write the formula for the angular position of dark fringes

The formula for the angular position of the dark fringes in a single-slit diffraction pattern is given by: $$ \sin{\theta} = (m + \frac{1}{2})\frac{\lambda}{w} $$ Where: \(\sin \theta\) is the sine of the angle of the dark fringe \(m\) is the order of the dark fringe (0, 1, 2, 3,...) \(\lambda\) is the wavelength of the light \(w\) is the width of the single slit We know that this problem involves the second dark fringe, which means \(m = 1\). We also know the width of the single slit, \(w = 2.00 \mathrm{~cm}\). Now, we will solve for \(\lambda\).
03

Solve for the wavelength of the light

Plug the given values and the angle we found in radians into the formula of the angular position of dark fringes: $$ \sin{\frac{43\pi}{180}} = \left(1 + \frac{1}{2}\right)\frac{\lambda}{2.00 \mathrm{~cm}} $$ To solve for \(\lambda\), first, multiply both sides by \(2.00 \mathrm{~cm}\): $$ (2.00 \mathrm{~cm})\sin{\frac{43\pi}{180}} = (1 + \frac{1}{2})\lambda $$ Now, we divide both sides by \((1 + \frac{1}{2})\): $$ \lambda = \frac{(2.00 \mathrm{~cm})\sin(\frac{43\pi}{180})}{1+\frac{1}{2}} $$ Finally, plug in the known values and compute for \(\lambda\): $$ \lambda = \frac{(2.00 \mathrm{~cm})\sin(\frac{43\pi}{180})}{1.5} \approx 7.18 \times 10^{-7} \mathrm{~m} $$
04

State the final answer

The wavelength of the light causing the second dark fringe at \(43.0^{\circ}\) in a single-slit diffraction pattern with a slit width of \(2.00 \mathrm{~cm}\) is approximately \(7.18 \times 10^{-7} \mathrm{~m}\).

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