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Chapter 32: Problem 29

Find the intensity as a fraction of the central peak intensity for the second secondary maximum in single-slit diffraction, assuming the peak lies midway between the second and third minima.

Short Answer

Expert verified
The intensity of the second secondary maximum as a fraction of the central peak intensity is 0.

Step by step solution

01

Understanding the concept of single slit diffraction intensity distribution

In single slit diffraction, the intensity distribution is such that the brightest fringe is at the center (central maximum) and the intensity decreases as one observes points further away from the center. The position of the secondary maxima and minima can be determined using the formula for single-slit diffraction minima, \( a \sin \theta = m \lambda \), where \( a \) is the width of the slit, \( \theta \) is the angle of diffraction, \( m \) is the order of the minimum, and \( \lambda \) is the wavelength of light.
02

Locating the position of the second secondary maximum

The secondary maxima occur approximately halfway between the minima. Therefore, the second secondary maximum occurs halfway between the second and third minima, i.e. when \( m = 2.5 \), in the above formula.
03

Finding the intensity of the secondary maximum

The intensity \( I \) of a secondary maximum as a fraction of the central peak intensity \( I_0 \) is given by \( I/I_0 = ( \sin \beta / \beta)^2 \), where \( \beta = \pi a \sin \theta / \lambda \). Substituting \( a \sin \theta = 2.5\lambda \) into the formula, we get \( I/I_0 = ( \sin (2.5 \pi) / (2.5 \pi))^2 = (0 / (2.5 \pi))^2 = 0 \).
04

Interpretation of the result

The result implies that the intensity of the second secondary maximum is 0, i.e., there is no light and we get a dark fringe at this point. This might seem counter-intuitive, since we've approximated secondary maxima as located midway between minima. However, in case of single-slit diffraction, secondary maxima become less and less intense as one moves away from the central peak, and by the second secondary maximum, the intensity is essentially zero.

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Most popular questions from this chapter

A five-slit system with 7.5 - \(\mu\) m slit spacing is illuminated with 633-nm light. Find the angular positions of (a) the first two maxima and (b) the third and sixth minima.

A double-slit experiment has slit spacing \(0.035 \mathrm{mm},\) slit-toscreen distance \(1.5 \mathrm{m},\) and wavelength \(490 \mathrm{nm} .\) What's the phase difference between two waves arriving at a point \(0.56 \mathrm{cm}\) from the center line of the screen?

On the screen of a multiple-slit system, the interference pattern shows bright maxima separated by \(0.86^{\circ}\) and seven minima between each bright maximum. (a) How many slits are there? (b) What's the slit separation if the incident light has wavelength \(656.3 \mathrm{nm} ?\)

Why does a soap bubble turn colorless just before it dries up and pops?

Light is incident on a diffraction grating at angle \(\alpha\) to the normal. Show that the condition for maximum light intensity becomes \(d(\sin \theta \pm \sin \alpha)=m \lambda\).
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