/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Light incident on a pair of slit... [FREE SOLUTION] | 91影视

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Chapter 25: Problem 31

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?]

Short Answer

Expert verified
Answer: The wavelength of the light is approximately 456 nm.

Step by step solution

01

List down given values

: Let's first write down the given values: - Distance between the screen and the slits (L) is \(2.50 \mathrm{m}\) - Slit separation (d) is \(0.0150 \mathrm{cm}\) - Distance between adjacent bright fringes (y) is \(0.760 \mathrm{cm}\) Now, let's convert the distances into the same unit (preferably meters). The slit separation, d = \(0.0150 \mathrm{cm} \times 10^{-2} = 1.50 \times 10^{-4} \mathrm{m}\) and the distance between adjacent bright fringes, y = \(0.760 \mathrm{cm} \times 10^{-2} = 7.60 \times 10^{-3} \mathrm{m}\).
02

Validate whether the small-angle approximation is justified

: The small-angle approximation is justified if the angle 胃 is very small, which means \(\tan(\theta) \approx \sin(\theta)\). This approximation is valid if \(y<<L\). Comparing the given values, we have \(7.60 \times 10^{-3} \mathrm{m} << 2.50 \mathrm{m}\). Since \(y\) is much smaller than \(L\), the small-angle approximation is justified.
03

Apply the double slit interference formula and solve for the wavelength

: The formula for double slit interference is: \(\frac{y}{L} \approx m\frac{\lambda}{d}\) Where: - \(m\) is the order of the bright fringe (an integer), considering two adjacent bright fringes, we can take \(m=1\), - \(\lambda\) is the wavelength of the light. Now, plug in the given values and solve for \(\lambda\): \(\lambda = \frac{y \times d}{L}\) \(\lambda = \frac{(7.60 \times 10^{-3} \mathrm{m}) \times (1.50 \times 10^{-4} \mathrm{m})}{2.50 \mathrm{m}}\) \(\lambda \approx 4.56 \times 10^{-7} \mathrm{m}\) Now, convert the wavelength back to nanometers (nm): \(\lambda \approx 4.56 \times 10^{-7} \mathrm{m} \times 10^9 = 456 \mathrm{nm}\) The wavelength of the light is around \(456 \mathrm{nm}\).

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