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A collimated beam of mercury green light at \(546.1 \mathrm{nm}\) is normally incident on a slit \(0.015 \mathrm{cm}\) wide. A lens of focal length \(60 \mathrm{cm}\) is placed behind the slit. A diffraction pattern is formed on a screen placed in the focal plane of the lens. Determine the distance between (a) the central maximum and first minimum and (b) the first and second minima.

Short Answer

Expert verified
(a) 0.00218 cm, (b) 0.00218 cm.

Step by step solution

01

Understand the problem

The problem involves a diffraction pattern created by a slit and a lens system. We need to find: (a) the distance from the central maximum to the first minimum and (b) the distance between the first and second minima on a screen located in the focal plane of the lens.
02

Identify relevant formulas

For diffraction minima from a single slit, the condition is given by \( a \sin \theta = m\lambda \), where \( a \) is the slit width, \( \theta \) is the angle of diffraction, \( \lambda \) is the wavelength, and \( m \) is the order of the minimum. The position \( y \) on the screen is given by \( y = f \tan \theta \), where \( f \) is the focal length of the lens.
03

Calculate angle for first minimum

For the first minimum, \( m = 1 \). Using \( a \sin \theta = m\lambda \), we have \( 0.015 \sin \theta = 1 \times 546.1\times 10^{-7} \). Solving for \( \sin \theta \), we find \( \sin \theta = \frac{546.1 \times 10^{-7}}{0.015} \approx 3.6407 \times 10^{-5} \).
04

Calculate position for first minimum

Assuming \( \theta \) is small, \( \tan \theta \approx \sin \theta \), thus \( y_1 = 60 \times 3.6407 \times 10^{-5} \approx 2.184 \times 10^{-3} \text{ cm} \). The distance from the central maximum to the first minimum \( y_1 \approx 0.002184 \text{ cm} \).
05

Calculate angle and position for second minimum

For the second minimum, \( m = 2 \). Using \( a \sin \theta = m \lambda \), solve \( 0.015 \sin \theta = 2 \times 546.1 \times 10^{-7} \). \( \sin \theta \approx 7.2813 \times 10^{-5} \), so \( y_2 = 60 \times 7.2813 \times 10^{-5} \approx 4.368 \times 10^{-3} \text{ cm} \).
06

Calculate distance between first and second minima

The distance between the first and second minima is \( y_2 - y_1 = 4.368 \times 10^{-3} - 2.184 \times 10^{-3} = 2.184 \times 10^{-3} \text{ cm} \).

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Key Concepts

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

Diffraction Pattern
When light encounters an obstacle or a slit that is comparable in size to its wavelength, it bends around the edges forming a unique light pattern on a screen. This phenomenon is called diffraction. When we project diffracted light from a single slit onto a screen, a distinct pattern emerges comprising a bright spot at the center known as the central maximum, and several less intense, spaced out bright and dark areas known as minima and maxima respectively.
The formation of such patterns arises due to the interference of light waves emanating through different parts of the slit. The spacing and intensity of the patterns depend heavily on factors like the wavelength of light and the slit width, indicating how a small change in any of these can dramatically alter the viewable diffraction pattern on the screen.
Central Maximum
The central maximum is the brightest point in a single slit diffraction pattern. It appears as the peak in the middle of the pattern and is the result of constructive interference of light rays passing directly through the slit at normal incidence. This region gets more light intensity compared to the neighboring areas because all paths through the slit contribute to the central point in-phase at the center of the screen.
If you imagine slicing the light waves into parts, each part would align perfectly when they meet in this central zone, causing the brightness to peak here. The central maximum is always at zero degrees angle and typically has the highest intensity.
First Minimum
Moving outward from the central maximum, the first point where the light intensity drops to zero is known as the first minimum. This occurs because the light rays from different parts of the slit interfere destructively at this angle, effectively canceling each other out. For a single slit, this is mathematically represented by the condition \( a \sin \theta = \lambda \), where \( a \) is the slit width, \( \theta \) is the angular position of the first minimum, and \( \lambda \) is the wavelength of the light.
Calculating the position on a screen, especially with a lens in place, involves transforming this angular advice into a linear distance using \( y = f \tan \theta \). For small angles, \( \tan \theta \approx \sin \theta \), thus giving a manageable approximation to find this first dim spot.
Second Minimum
The second minimum is further from the central maximum and is the next point of zero light intensity beyond the first minimum. It is the result of continued destructive interference at a larger angle, effectively twice that of the first minimum. The mathematical description follows the condition \( a \sin \theta = 2\lambda \), using the same symbols as for the first minimum.
On the screen, as light travels further, the disrupted light paths continue to cancel each other out precisely at these minima, forming dark spots. Calculating the distance between sequential minima (such as from the first to the second) becomes a geometric progression using the above conditions, which determines not only the pattern but also spacing on the screen.

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Most popular questions from this chapter

A telescope objective is \(12 \mathrm{cm}\) in diameter and has a focal Iength of \(150 \mathrm{cm} .\) Light of mean wavelength \(550 \mathrm{nm}\) from a distant star enters the scope as a nearly collimated beam. Compute the radius of the central disk of light forming the image of the star on the focal plane of the lens.

a. Show that the number of bright fringes seen under the central diffraction peak in a Fraunhofer double-slit pattern is given by \(2(a / b)-1,\) where \(a / b\) is the ratio of slit separation to slit width. b. If 13 bright fringes are seen in the central diffraction peak when the slit width is \(0.30 \mathrm{mm}\), determine the slit separation.

A double-slit diffraction pattern is formed using mercury green light at \(546.1 \mathrm{nm}\). Each slit has a width of \(0.100 \mathrm{mm}\) The pattern reveals that the fourth-order interference maxima are missing from the pattern. a. What is the slit separation? b. What is the irradiance of the first three orders of interference fringes, relative to the zeroth-order maximum?

The width of a rectangular slit is measured in the laboratory by means of its diffraction pattern at a distance of \(2 \mathrm{m}\) from the slit. When illuminated normally with a parallel beam of laser light \((632.8 \mathrm{nm}),\) the distance between the third minima on either side of the principal maximum is measured. An average of several tries gives \(5.625 \mathrm{cm}\) a. Assuming Fraunhofer diffraction, what is the slit width? b. Is the assumption of far-field diffraction justified in this case? What is the ratio \(L / L_{\min } ?\)

Assume that a 2-mm-diameter laser beam (632.8 nm) is diffraction limited and has a constant irradiance over its cross section. On the basis of spreading due to diffraction alone, how far must it travel to double its diameter?

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