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Light of wavelength \(5.30 \times 10^{2} \mathrm{~nm}\) illuminates a pair of slits separated by \(0.300 \mathrm{~mm}\). If a screen is placed \(2.00 \mathrm{~m}\) from the slits, determine the distance between the first and second dark fringes.

Short Answer

Expert verified
The final answer will be the difference between the first and second dark fringe distances calculated in Step 4. It is important to convert all given measurements in the same unit system before performing the calculations.

Step by step solution

01

Identify Knowns and Unknowns

The known variables in the problem are the wavelength of light (\(\lambda\) = 530 nm), the distance between slits (d = 0.300 mm) and the distance from slits to the screen (L = 2.00m). The unknown we are solving for is the distance between the first and second dark fringes.
02

Use Formula for Dark Fringes to Determine Angular Position

For dark fringes in a double-slit diffraction pattern, the following formula applies: \(d sin(\Theta) = m\lambda\). In this case, d is the distance between the slits, \(\Theta\) is the angular position, m is the order of the dark fringe, and \(\lambda\) is the light wavelength.
03

Compute Angular Positions

Now let's calculate the angular positions of the first (m=1) and second (m=2) dark fringes using the aforementioned equation.
04

Convert Angular Positions to Distances

For small angles (which will be the case here), the distance of a point from the center of the pattern on the screen (y) can be approximated as \(L\tan(\Theta)\), where L is the distance between the slits and the screen. The distance between the first and second dark fringes will be the difference between the two fringe distances we compute.

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Key Concepts

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

Wave Interference
Wave interference is a fundamental concept in physics that describes how two or more waves superpose to form a resultant wave. This phenomenon occurs when waves overlap, and it can lead to an increase (constructive interference) or decrease (destructive interference) in wave amplitude. In the context of light, a wave-like form of energy, interference results in patterns of bright and dark fringes.

This can be observed in experiments such as the double-slit experiment, where light waves pass through two narrow slits and interfere with one another on the other side, creating an interference pattern on a screen. The areas where the waves reinforce each other appear as bright bands, and the areas where they cancel out appear as dark bands. Understanding this concept is crucial for analyzing diffraction and interference patterns in exercises like the one provided.
Fringe Pattern Calculation
The calculation of fringe patterns is an integral part of understanding wave interference in double-slit experiments. To determine the distance between fringes, one must know the formula that relates the slit separation, the wavelength of light, and the distance from the slits to the screen. For dark fringes, the formula is given by \[\begin{equation}d \sin(\Theta) = m\lambda\end{equation}\]where \(d\) is the distance between slits, \(\Theta\) is the angular position, \(m\) is the order of dark fringe (integer values, where \(m = 1, 2, 3, ...\) represent the first, second, third, etc., dark fringes), and \(\lambda\) is the wavelength of light.

When analyzing the problem provided, we first calculate the angular positions of the first and second dark fringes using this formula and then use the small-angle approximation (\(\tan(\Theta) \approx \sin(\Theta)\)) to convert these angles into distances on the screen. By finding the distances for both the first and second dark fringes and taking the difference, we obtain the distance between the two fringes.
Diffraction and Interference
Diffraction and interference are two phenomena that often occur together, especially in the study of light and wave behavior. Diffraction refers to the bending and spreading out of waves as they pass around an obstacle or through an opening. It's what allows waves to 'turn corners' and propagate beyond the geometrical shadow of the obstruction.

Interference, on the other hand, relates to the combination of two or more waves overlapping and forming a new wave pattern. In the double-slit experiment, diffraction occurs as the light waves pass through the slits and subsequently interfere with one another to produce bright and dark fringes on a screen.

Since these phenomena are closely related, their combined effect must be considered to accurately predict the resulting light pattern observed in such experiments, as exemplified in the problem at hand where understanding both has enabled us to calculate the distance between dark fringes on the screen.

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Most popular questions from this chapter

Take red light at \(700 \mathrm{~nm}\) and violet at \(400 \mathrm{~nm}\) as the ends of the visible spectrum and consider the continuous spectrum of white light formed by a diffraction grating with a spacing of \(d\) meters between adjacent lines. Show that the interval \(\theta_{i 2} \leq \theta \leq \theta_{r 2}\) of the continuous spectrum in second order must overlap the interval \(\theta_{\mathrm{m} 3} \leq \theta \leq \theta_{\mathrm{r} 3} \mathrm{of}\) the third-order spectrum. Note: \(\theta_{n / 1}\) is the angle of the violet light in second order, and \(\theta_{r 2}\) is the angle made by red light in second order.

Sunlight is incident on a diffraction grating that has 2750 lines \(/ \mathrm{cm}\). The second-order spectrum over the visible range \((400-700 \mathrm{~nm})\) is to be limited to \(1.75 \mathrm{~cm}\) along a screen that is a distance \(L\) from the grating. What is the required value of \(L\) ?

Plane-polarized light is incident on a single polarizing disk, with the direction of \(E_{0}\) parallel to the direction of the transmission axis. Through what angle should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of (a) \(2.00\), (b) \(4.00\), and (c) \(6.00 ?\)

A thin film of \(\mathrm{MgF}_{2}(n=1.38)\) with thickness \(1.00 \times\) \(10^{-5} \mathrm{~cm}\) is used to coat a camera lens. Are any wave- lengths in the visible spectrum intensified in the reflected light?

Light of wavelength \(600 \mathrm{~nm}\) falls on a \(0.40\) -mm-wide slit and forms a diffraction pattern on a screen \(1.5 \mathrm{~m}\) away. (a) Find the position of the first dark band on each side of the central maximum. (b) Find the width of the central maximum.

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