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Chapter 4: Problem 91

A source of light having two wavelengths \(550 \mathrm{nm}\) and \(600 \mathrm{nm}\) of equal intensity is incident on a slit of width \(1.8 \mu \mathrm{m} .\) Find the separation of the \(m=1\) bright spots of the two wavelengths on a screen \(30.0 \mathrm{cm}\) away.

Short Answer

Expert verified
The separation between the m = 1 bright spots of the two wavelengths on the screen is approximately 8.7 mm.

Step by step solution

01

Write down the given information.

We are given the following information: - Wavelength of first light source (λ1) = 550 nm - Wavelength of second light source (λ2) = 600 nm - Width of the slit (a) = 1.8 μm - Distance from the slit to the screen (L) = 30 cm First, we need to convert all the given units to meters: - λ1 = 550 * 10^(-9) m - λ2 = 600 * 10^(-9) m - a = 1.8 * 10^(-6) m - L = 0.3 m
02

Apply the single-slit diffraction formula.

The single-slit diffraction formula for the first-order bright spots (m = 1) is given as: \(sin(\theta) \approx \frac{m \lambda}{a}\) For the first light source, let's call the angle θ1: \(sin(\theta_1) \approx \frac{1 \cdot (550 * 10^{-9})}{1.8 * 10^{-6}}\) For the second light source, let's call the angle θ2: \(sin(\theta_2) \approx \frac{1 \cdot (600 * 10^{-9})}{1.8 * 10^{-6}}\)
03

Solve for the angles θ1 and θ2.

Calculating the angles for both light sources: \(sin(\theta_1) \approx 0.3056\) Therefore, \(\theta_1 \approx 17.9°\) \(sin(\theta_2) \approx 0.3333\) Therefore, \(\theta_2 \approx 19.5°\)
04

Calculate the separation between the bright spots of both wavelengths.

Using the small-angle approximation, we can now calculate the separation between the first-order bright spots for the two wavelengths (Δy) using the formula: \(\Delta y = L \cdot tan(\theta_2 - \theta_1)\) Plugging in the values, we get: \(\Delta y = 0.3 \cdot tan(19.5° - 17.9°)\) \(\Delta y \approx 0.0087 \: m\)
05

Convert the separation to the required unit and write the final answer.

The separation between the first-order bright spots of the two wavelengths is 0.0087 meters. To convert it to millimeters, multiply by 1000: \(\Delta y \approx 8.7 \: mm\) Thus, the separation between the m = 1 bright spots of the two wavelengths on the screen is approximately 8.7 mm.

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Most popular questions from this chapter

A diffraction grating produces a second maximum that is \(89.7 \mathrm{cm}\) from the central maximum on a screen \(2.0 \mathrm{m}\) away. If the grating has 600 lines per centimeter, what is the wavelength of the light that produces the diffraction pattern?

A light ray of wavelength 461.9 nm emerges from a 2 -mm circular aperture of a krypton ion laser. Due to diffraction, the beam expands as it moves out. How large is the central bright spot at (a) \(1 \mathrm{m},\) (b) \(1 \mathrm{km},\) (c) \(1000 \mathrm{km}\) and (d) at the surface of the moon at a distance of 400,000 km from Earth.

How far apart must two objects be on the moon to be distinguishable by eye if only the diffraction effects of the eye's pupil limit the resolution? Assume \(550 \mathrm{nm}\) for the wavelength of light, the pupil diameter \(5.0 \mathrm{mm}\), and \(400,000 \mathrm{km}\) for the distance to the moon.

As an example of diffraction by apertures of everyday dimensions, consider a doorway of width \(1.0 \mathrm{m}\). (a) What is the angular position of the first minimum in the diffraction pattem of 600 -nm light? (b) Repeat this calculation for a musical note of frequency \(440 \mathrm{Hz}\) (A above middle C). Take the speed of sound to be \(343 \mathrm{m} / \mathrm{s}\).

Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular spreading of a flashlight beam that is originally \(5.00 \mathrm{cm}\) in diameter with an average wavelength of \(600 \mathrm{nm}\)
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