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Parallel rays of light with wavelength 620 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of 40.0 \(\mathrm{cm} .\) The diffraction pattern is observed in the focal plane of the lens, and the distance from the center of the central maximum to the first minimum is 36.5 \(\mathrm{cm} .\) What is the width of the slit? (Note: The angle that locates the first minimum is not small.)

Short Answer

Expert verified
The width of the slit is 922 nm or 9.22 x 10^-5 cm.

Step by step solution

01

Understand the Problem

We are given a situation where light with a wavelength of 620 nm passes through a slit and a diffraction pattern is formed on a screen placed in the focal plane of a lens. The focal length of the lens is 40.0 cm, and the distance to the first minimum from the central maximum is 36.5 cm. We need to find the width of the slit.
02

Recall the Diffraction Formula

For a single slit diffraction pattern, the position of the first minimum is given by the formula: \( a \sin \theta = m \lambda \), where \( a \) is the width of the slit, \( \theta \) is the angle of the first minimum, \( m \) is the order of the minimum (in this case, \( m = 1 \)), and \( \lambda \) is the wavelength of the light. We need to find the slit width \( a \).
03

Use the Lens Formula

Using the geometry of the setup, \( \tan \theta = \frac{x}{f} \), where \( x \) is the distance from the central maximum to the first minimum (36.5 cm) and \( f \) is the focal length of the lens (40.0 cm). Since the angle \( \theta \) is not small, we cannot use the small angle approximation.
04

Calculate \( \theta \) using \( \tan \theta \)

Calculate \( \theta \) as follows: \( \tan \theta = \frac{36.5}{40} = 0.9125 \). Find \( \theta \) by taking the arctan: \( \theta = \arctan(0.9125) \). Using a calculator, we find \( \theta \approx 42.19^\circ \).
05

Solve for Slit Width \( a \)

Use the diffraction formula: \( a \sin \theta = \lambda \). First calculate \( \sin \theta \): \( \sin(42.19^\circ) \approx 0.6717 \). Substitute this value and the wavelength into the equation: \( a \cdot 0.6717 = 620 \times 10^{-9} \ m \). Solve for \( a \): \( a = \frac{620 \times 10^{-9}}{0.6717} \approx 922 \times 10^{-9} \ m = 922 \ nm \).
06

Convert Slit Width to Centimeters

Convert the slit width from nanometers to centimeters for clarity in the answer: \( 922 \ nm = 9.22 \times 10^{-5} \ cm \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Single Slit Diffraction
When light waves pass through a narrow opening like a slit, they spread out or "bend," creating a pattern of light and dark bands known as a diffraction pattern. This phenomenon occurs due to interference between the light waves. This is called **single slit diffraction** because the light is passing through one slit. In this case, the light waves overlap and interfere with each other, either enhancing or canceling out the light intensity at different points. The central bright band, or central maximum, is flanked by alternating dark and bright bands. The position of these minima (dark bands) can be calculated, and they hold vital information about the light's wavelength and the slit width. The location of the first minimum is particularly significant because it can help determine the slit width using the diffraction formula. Understanding how light behaves in single slit diffraction is crucial as it applies to various scientific and engineering applications, such as optics and photography.
Wavelength
**Wavelength** is the distance between consecutive peaks of a wave. When you think about light, it defines the color of the light we see. In this exercise, the light has a wavelength of 620 nm (nanometers). Wavelength is crucial in diffraction because it affects how much a wave spreads out after passing a slit. Shorter wavelengths tend to spread less than longer ones. In equations and calculations, wavelength is denoted by the Greek letter lambda (\(\lambda\)).Remember that visible light ranges from about 400 nm (violet) to about 700 nm (red). So 620 nm is in the red-orange portion of the spectrum. The wavelength, along with the slit width and angle, helps determine the pattern's appearance.
Focal Length
The **focal length** of a lens is the distance between the center of the lens and the point where it focuses parallel beams of light. In this problem, the focal length is 40 cm. This lens helps to project the diffraction pattern onto a screen, making it easier to analyze and measure. The lens acts as a tool that aids in the observation of the diffraction pattern. The focal plane, where the pattern forms, lies exactly at this focal length from the lens. It’s important because the distance influences how the light waves spread out. In a practical sense, lenses are used to capture images in cameras, in microscopes to magnify small objects, and even in our eyes.
Diffraction Pattern
The **diffraction pattern** formed when light passes through a single slit consists of several dark and bright bands. The central maximum is the bright band at the center and is the widest. First minima are the dark bands on either side of the central maximum, where the light intensity drops to zero. This happens because of interference: at certain angles, the light waves coming from different parts of the slit cancel each other out (destructive interference) resulting in minima. Where the waves add up (constructive interference), we see maxima. Knowing the distance from the central maximum to a minimum, and using the single slit diffraction formula, one can calculate the width of the slit. The pattern itself is tangible proof of the wave nature of light, showcasing experiments and derivations leading to our fundamental understanding of waves.

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Most popular questions from this chapter

Identifying Isotones hy Snectra Different isotones of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is 656.45 nm; for deuterium, the corresponding wavelength is 656.27 \(\mathrm{nm} .\) (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda=587.9782 \mathrm{nm}\) and \(\lambda=\) 587.8002 \(\mathrm{nm} .\) You wish to resolve these spectral lines in first order using a grating 1.20 \(\mathrm{cm}\) in length. What minimum number of slits per centimeter must the grating have?

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

Quasars, an abbreviation for quasi-stellar radio sources, are distant objects that look like stars through a telescope but that emit far more electromagnetic radiation than an entire normal galaxy of stars. An example is the bright object below and to the left of center in Fig. P36.66; the other elongated objects in this image are normal galaxies. The leading model for the structure of a quasar is a galaxy with a supermassive black hole at its center. In this model, the radiation is emitted by interstellar gas and dust within the galaxy as this material falls toward the black hole. The radiation is thought to emanate from a region just a few light-years radiation is thought to emanate from a region just a few light-years in diameter. (The diffuse glow surrounding the bright quasar shown in Fig. \(P 36.66\) is thought to be this quasar's host galaxy.) To investigate this model of quasars and to study other exotic astronomical objects, the Russian Space Agency plans to place a radio telescope in an orbit that extends to \(77,000 \mathrm{km}\) from the earth. When the signals from this telescope are combined with signals from the ground-based telescopes of the VLBA, the resolution will be that of a single radio telescope \(77,000 \mathrm{km}\) in diameter. What is the size of the smallest detail that this arrangement could resolve ir quasar \(3 \mathrm{C} 405,\) which is \(7.2 \times 10^{8}\) light-years from earth, using radio waves at a frequency of 1665 \(\mathrm{MHz}\) ? (Hint: Use Rayleigh's criterion.) Give your answer in light-years and in kilometers.

A diffraction grating has 650 slits/mm. What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm.}\) )

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