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(a) If the wavelength used in a double-slit experiment is decreased, the distance between adjacent maxima will (1) increase, (2) decrease, (3) remain the same. Explain. (b) If the separation between the two slits is \(0.20 \mathrm{~mm}\) and the adjacent maxima of the interference pattern on a screen \(1.5 \mathrm{~m}\) away from the slits are \(0.45 \mathrm{~cm}\) apart, what is the wavelength and color of the light? (c) If the wavelength is \(550 \mathrm{nm}\), what is the distance between adjacent maxima?

Short Answer

Expert verified
(a) Decrease. (b) 600 nm, orange. (c) 0.0733 cm.

Step by step solution

01

Understanding the Double-Slit Formula

In a double-slit experiment, the distance between adjacent maxima (bright fringes) on the screen is given by the formula: \( x = \frac{\lambda D}{d} \), where \( x \) is the fringe spacing, \( \lambda \) is the wavelength of the light, \( D \) is the distance from the slits to the screen, and \( d \) is the separation between the slits.
02

Effect of Decreasing Wavelength

(a) According to the formula \( x = \frac{\lambda D}{d} \), decreasing the wavelength \( \lambda \) results in a decrease in \( x \), the distance between adjacent maxima. Therefore, the distance between adjacent maxima will decrease.
03

Rearrange the Formula to Solve for Wavelength

(b) Using the given values for slit separation \( d = 0.20 \text{ mm} = 0.0002 \text{ m} \), screen distance \( D = 1.5 \text{ m} \), and fringe spacing \( x = 0.45 \text{ cm} = 0.0045 \text{ m} \), rearrange the formula for \( \lambda \):\[ \lambda = \frac{xd}{D} \]
04

Calculate the Wavelength

Substitute the known values into the rearranged formula: \( \lambda = \frac{0.0045 \times 0.0002}{1.5} \), resulting in \( \lambda = 6.0 \times 10^{-7} \text{ m} = 600 \text{ nm} \). The wavelength corresponds to orange light.
05

Calculate New Fringe Spacing for Given Wavelength

(c) If \( \lambda = 550 \text{ nm} = 5.5 \times 10^{-7} \text{ m} \), using the same \( d \) and \( D \), compute \( x \) as follows: \[ x = \frac{(5.5 \times 10^{-7}) \times 0.0002}{1.5} \], which gives \( x = 0.0000733 \text{ m} = 0.0733 \text{ mm} \) or \( 0.0733 \text{ cm} \).

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Key Concepts

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

Wavelength
Wavelength is a key factor in understanding the behavior of waves, particularly in the context of a double-slit experiment. It refers to the distance between consecutive points of equal phase, such as adjacent crests, in a wave. In the double-slit experiment, the wavelength (\(\lambda\)) of light plays a crucial role in determining the position and spacing of the interference pattern seen on a screen.
In this experiment, if the wavelength is decreased, the spacing between adjacent bright fringes (maxima) will decrease. This is because wavelength is directly proportional to fringe spacing (\(x\)) in the formula:
\[ x = \frac{\lambda D}{d} \]
Here:
  • \(\lambda\) is the wavelength of the light.
  • \(D\) is the distance from the slits to the screen.
  • \(d\) is the separation between the slits.
Decreasing the wavelength therefore makes the interference pattern more closely packed on the projection screen.
Interference Pattern
The interference pattern in a double-slit experiment is a visual representation of wave interference. When light waves pass through two slits, they overlap and intersect on the other side, creating a pattern of light and dark areas due to constructive and destructive interference.
Constructive interference occurs when the crests and troughs of the light waves line up perfectly, amplifying the light's intensity and forming bright fringes (maxima). Destructive interference happens when crests meet troughs, canceling each other out and resulting in darkness (minima).
The regular, periodic pattern of light and dark bands is what we refer to as the interference pattern. It can be used to measure the wavelength of light, as any change in the wavelength will affect the spacing and position of these fringes on the screen. The predictability of this pattern demonstrates the wave nature of light.
Fringe Spacing
Fringe spacing in the context of a double-slit experiment refers to the distance between consecutive bright fringes on the screen. This spacing is crucial because it allows us to link the physical setup of the experiment with observable phenomena.
Using the formula, the fringe spacing \(x\) can be determined by:
\[ x = \frac{\lambda D}{d} \]
From this formula, we can see:
  • Fringe spacing \(x\) increases with a larger wavelength \(\lambda\).
  • Fringe spacing \(x\) increases with a larger distance to the screen \(D\).
  • Fringe spacing \(x\) decreases with a greater separation between the slits \(d\).
In experimental setups, measuring fringe spacing allows scientists to backtrack and determine the wavelength of the incoming light based on the setup parameters. This demonstrates the interconnectedness of wave phenomena.
Light Color
The color of light is directly linked to its wavelength. In visible light, different colors appear due to varying wavelengths.
  • Violet light has the shortest wavelength, around 400 nm.
  • Red light has the longest wavelength, around 700 nm.
  • Other colors fall in between, like blue, green, yellow, and orange.
In the step-by-step solution provided, the calculated wavelength was 600 nm, which corresponds to orange light. By determining the wavelength, we can infer the color of the light used in the experiment.
Therefore, understanding how wavelength impacts both the appearance and properties of light offers a hands-on application of these concepts in optics and beyond. This is essential in fields such as telecommunications, imaging, and even in everyday technology like smartphones and TV displays.

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Most popular questions from this chapter

A lens with an index of refraction of 1.60 is to be coated with a material \((n=1.40)\) that will make the lens nonreflecting for yellow-orange light \((\lambda=515 \mathrm{nm})\) normally incident on the lens. What is the minimum required thickness of the coating?

Find the angles of the blue \((\lambda=420 \mathrm{nm})\) and red \((\lambda=680 \mathrm{nm})\) components of the first- and second-order maxima in a pattern produced by a diffraction grating with 7500 lines \(/ \mathrm{cm}\).

Light is incident on a certain material in air. (a) If the index of refraction of the material increases, the polarizing (Brewster) angle will (1) also increase, (2) decrease, (3) remain the same. Explain. (b) What are the polarizing angles if the index of refraction is 1.6 and \(1.8 ?\)

The angle of incidence is adjusted so there is maximum linear polarization for the reflected light from a transparent piece of plastic in air. (a) There is (1) no, (2) maximum, or (3) some light transmitted through the plastic. Explain. (b) If the index of refraction of the plastic is 1.40 , what would be the angle of refraction in the plastic?

A thin air wedge between two flat glass plates forms bright and dark interference bands when illuminated with normally incident monochromatic light. (See Fig. 24.9.) (a) Show that the thickness of the air wedge changes by \(\lambda / 2\) from one bright band to the next, where \(\lambda\) is the wavelength of the light. (b) What would be the change in the thickness of the wedge between bright bands if the space were filled with a liquid with an index of refraction \(n ?\)

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