Monochromatic light of wavelength 530 nm passes through a horizontal single slit of width \(1.5 \mu \mathrm{m}\) in an opaque plate. A screen of dimensions \(2.0 \mathrm{m} \times 2.0 \mathrm{m}\) is \(1.2 \mathrm{m}\) away from the slit. (a) Which way is the diffraction pattern spread out on the screen? (b) What are the angles of the minima with respect to the center? (c) What are the angles of the maxima? (d) How wide is the central bright fringe on the screen? (e) How wide is the next bright fringe on the screen?

Short Answer

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(a) The diffraction pattern will spread horizontally on the screen. (b) The angles of the minima, \(\theta\), can be calculated using: \(\theta = \arcsin{\frac{m \lambda}{w}}\). (c) The angles of the maxima, \(\theta\), can be calculated using: \(\theta = \arcsin{\frac{(m +\frac{1}{2}) \lambda}{w}}\). (d) The width of the central bright fringe is \(2y_1\), where \(y_1 = L\tan{\theta_1}\). (e) The width of the next bright fringe is \(2(y_2 - y_1)\), where \(y_2 = L\tan{\theta_2}\).

Step by step solution

01

(a) Determine the direction of the diffraction pattern spread

To identify which way the diffraction pattern spreads, we simply need to understand that the diffraction pattern will spread perpendicularly from the slit. Therefore, the pattern will spread horizontally on the screen.
02

(b) Calculate the angles of the minima

To find the angles of the minima, we can use the formula for single-slit diffraction minima: \(m \lambda = w \sin{\theta}\) Where \(m\) is the order of the minimum, \(\lambda\) is the wavelength of the light, \(w\) is the width of the slit, and \(\theta\) is the angle between the center of the pattern and the minimum. We can rewrite the formula for \(\theta\): \(\theta = \arcsin{\frac{m \lambda}{w}}\) For m = 1, 2, 3,... we can calculate the respective angles of the minima using the given values of \(\lambda\) and \(w\).
03

(c) Calculate the angles of the maxima

To find the angles of the maxima, we can use the formula for single-slit diffraction maxima: \((m +\frac{1}{2})\lambda = w \sin{\theta}\) Where \(m\) is the non-negative integer for the maxima (0, 1, 2, ...), \(\lambda\) is the wavelength, \(w\) is the width of the slit, and \(\theta\) is the angle between the center of the pattern and the maxima. Rewriting the formula for \(\theta\): \(\theta = \arcsin{\frac{(m +\frac{1}{2}) \lambda}{w}}\) For m = 0, 1, 2,... we can calculate the respective angles of the maxima using the given values of \(\lambda\) and \(w\).
04

(d) Calculate the width of the central bright fringe

The central bright fringe extends from the central maximum to the first minimum on either side. We can calculate the distance on the screen from the center to the first minimum by using trigonometry: \(y_1 = L\tan{\theta_1}\) Where \(y_1\) is the distance from the center to the first minimum on the screen, \(L\) is the distance between the slit and the screen, and \(\theta_1\) is the angle of the first minimum as calculated in part (b). Since the central bright fringe is symmetrical about the central maximum, its total width on the screen will be: Width of central bright fringe = \(2y_1\)
05

(e) Calculate the width of the next bright fringe

The next bright fringe is the region between the first and second minima. We can calculate the distance on the screen from the center to the second minimum by using trigonometry: \(y_2 = L\tan{\theta_2}\) Where \(y_2\) is the distance from the center to the second minimum on the screen, \(L\) is the distance between the slit and the screen, and \(\theta_2\) is the angle of the second minimum as calculated in part (b). Since the fringe is symmetrical about the central maximum, the width of the next bright fringe on the screen will be: Width of next bright fringe = \(2(y_2 - y_1)\)

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Key Concepts

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

Diffraction Pattern
When monochromatic light passes through a small slit, it doesn't simply travel in straight lines; instead, it bends and spreads out. This bending and spreading of light is known as diffraction, and the resulting band of light and dark areas is called the diffraction pattern. In our exercise, a horizontal single slit causes the light to form a diffraction pattern that spreads horizontally across the screen. To visualize this, imagine water waves hitting a narrow barrier with a slit; the waves spread out in a semi-circular pattern on the other side, similar to how light creates a diffraction pattern on the screen.

The diffraction pattern consists of a central bright fringe, where the light waves constructively interfere, surrounded by alternating dark and bright fringes. The dark areas are the result of destructive interference, where waves cancel each other out. Understanding the diffraction pattern helps us predict where light will be most intense and where it will be absent, which is essential for applications like optical instruments and engineering designs.
Angle of Minima
The angle of minima in a single-slit diffraction pattern represents the direction where the destructive interference occurs, creating dark bands on the screen. Mathematically, these angles can be found using the formula
\( m \lambda = w \sin{\theta} \)
where m is an integer (1, 2, 3,...) denoting the order of the minimum, \lambda is the wavelength of the light, w is the width of the slit, and \theta is the angle of minima with respect to the center of the diffraction pattern. By measuring these angles, or calculating them with the given wavelength and slit width, we can determine where the dark bands will appear on the screen, and this helps in predicting the overall shape and spacing of the diffraction pattern.
Angle of Maxima
Conversely, the angle of maxima reflects where in the diffraction pattern we expect to see the bright fringes due to constructive interference. The equation to find the angle of maxima is subtly different from that of the minima:
\((m +\frac{1}{2})\lambda = w \sin{\theta}\)
Here, m is again an integer (0, 1, 2,...), but for maxima, we add a half before multiplying by the wavelength. By calculating the angle of maxima, we can pinpoint the location of bright bands on the screen. These bright fringes are indicative of points where light waves reinforce each other, strengthening the overall intensity of the light.
Central Bright Fringe
The central bright fringe is the hallmark of the single-slit diffraction pattern, located at the center of the pattern and invariably the brightest and widest band. This band is created by the direct beams of light that pass through the slit without being deflected. It is symmetrical and is usually used as a reference point for measuring the position of other fringes. Its width directly influences the spacing of subsequent bright and dark bands. In our exercise, we determine this width by finding the distance from the center to the first minimum on either side of the central maximum, which helps us understand just how much the light spreads out and how it relates to the slit width and wavelength.

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Most popular questions from this chapter

(a) What is the minimum angular spread of a 633 -nm wavelength He-Ne laser beam that is originally \(1.00 \mathrm{mm}\) in diameter? (b) If this laser is aimed at a mountain cliff 15.0 km away, how big will the illuminated spot be? (c) How big a spot would be illuminated on the moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round- trip time and, hence, distance.)

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 \(3.00 \mu \mathrm{m}\)

(a) The dwarf planet Pluto and its moon, Charon, are separated by \(19,600 \mathrm{km}\). Neglecting atmospheric effects, should the 5.08 -m-diameter Palomar Mountain telescope be able to resolve these bodies when they are \(4.50 \times 10^{9} \mathrm{km} \quad\) from Earth? Assume an average wavelength of \(550 \mathrm{nm}\). (b) In actuality, it is just barely possible to discern that Pluto and Charon are separate bodies using a ground-based telescope. What are the reasons for this?

(a) If a single slit produces a first minimum at \(14.5^{\circ}\) at what angle is the second-order minimum? (b) What is the angle of the third-order minimum? (c) Is there a fourthorder minimum? (d) Use your answers to illustrate how the angular width of the central maximum is about twice the angular width of the next maximum (which is the angle between the first and second minima).

Two lamps producing light of wavelength 589 nm are fixed \(1.0 \mathrm{m}\) apart on a wooden plank. What is the maximum distance an observer can be and still resolve the lamps as two separate sources of light, if the resolution is affected solely by the diffraction of light entering the eye? Assume light enters the eye through a pupil of diameter \(4.5 \mathrm{mm}\).

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