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Chapter 4: Problem 21

Find the wavelength of light that has its third minimum at an angle of \(48.6^{\circ}\) when it falls on a single slit of width \(3.00 \mu \mathrm{m}\)

Short Answer

Expert verified
The wavelength of the light that has its third minimum at an angle of 48.6° when it falls on a single slit of width 3.00 μm is approximately \(2.14 \mu\mathrm{m}\).

Step by step solution

01

Identify the Given Information and the Target Variable

We are given: - The angle for the third minimum (dark fringe) is 48.6° - The slit width is 3.00 μm - Our target variable is the wavelength of the light
02

Write Down the Single-Slit Diffraction Formula

The single-slit diffraction formula relates the angle of a dark fringe minimum (θ), the slit width (a), the order of the minimum (m), and the wavelength of the light (λ) as follows: \[ a \sin(\theta) = m \lambda \]
03

Substitute the Given Information into the Formula

We are given the angle (θ = 48.6°), the slit width (a = 3.00 μm), and the order of the minimum (m = 3, since it's the third minimum). We'll substitute these values into the formula: \[ (3.00\, \mu\mathrm{m}) \sin(48.6^{\circ}) = 3 \lambda \]
04

Solve for the Wavelength of the Light

Rearrange the formula to solve for the wavelength (λ) by dividing both sides by the order of the minimum (m = 3): \[ \lambda = \frac{(3.00\, \mu\mathrm{m}) \sin(48.6^{\circ})}{3} \] Now, use a calculator to compute the value: \[ \lambda = \frac{(3.00\, \mu \mathrm{m}) \sin(48.6^{\circ})}{3} \approx 2.14 \mu\mathrm{m} \]
05

Report the Answer

The wavelength of the light that has its third minimum at an angle of 48.6° when it falls on a single slit of width 3.00 μm is approximately 2.14 μm.

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Most popular questions from this chapter

Radio telescopes are telescopes used for the detection of radio emission from space. Because radio waves have much longer wavelengths than visible light, the diameter of a radio telescope must be very large to provide good resolution. For example, the radio telescope in Penticton, BC in Canada, has a diameter of \(26 \mathrm{m}\) and can be operated at frequencies as high as \(6.6 \mathrm{GHz}\). (a) What is the wavelength corresponding to this frequency? (b) What is the angular separation of two radio sources that can be resolved by this telescope? (c) Compare the telescope's resolution with the angular size of the moon.

The width of the central peak in a single-slit diffraction pattern is \(5.0 \mathrm{mm}\). The wavelength of the light is \(600 \mathrm{nm}\), and the screen is \(2.0 \mathrm{m}\) from the slit. (a) What is the width of the slit? (b) Determine the ratio of the intensity at \(4.5 \mathrm{mm}\) from the center of the pattern to the intensity at the center.

Can an astronaut orbiting Earth in a satellite at a distance of \(180 \mathrm{km}\) from the surface distinguish two skyscrapers that are \(20 \mathrm{m}\) apart? Assume that the pupils of the astronaut's eyes have a diameter of \(5.0 \mathrm{mm}\) and that most of the light is centered around \(500 \mathrm{nm}\).

When a monochromatic light of wavelength 430 nm incident on a double slit of slit separation \(5 \mu \mathrm{m}\), there are 11 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength \(632.8 \mathrm{nm}\) for the same double slit?

(a) Show that a 30,000 line per centimeter grating will not produce a maximum for visible light. (b) What is the longest wavelength for which it does produce a firstorder maximum? (c) What is the greatest number of line per centimeter a diffraction grating can have and produce a complete second-order spectrum for visible light?
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