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Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 \(\mathrm{cm}\) from the slits. The bright bands are 1.00 \(\mathrm{cm}\) apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

Short Answer

Expert verified
Slit separation \( d = 45 \text{ µm} \); width \( a = 33.3 \text{ µm} \).

Step by step solution

01

Convert Units

First, convert the given wavelength from nanometers to meters: 500.0 nm = 500.0 x 10^-9 m. We also note that 90.0 cm = 0.9 m.
02

Understand the Interference Pattern

The interference pattern means that light is diffracted through slits, and these slits also cause a diffraction pattern due to their width. The bright bands' separation of 1.00 cm corresponds to the distance between adjacent maxima in the interference pattern.
03

Use the Formula for Bright Fringe Separation

The distance between adjacent bright fringes (y) in the interference pattern is given by \[ y = \frac{m \lambda L}{d} \] where \( m \) is the order number, \( \lambda \) the wavelength (500.0 x 10^-9 m), \( L \) the distance to the screen (0.9 m), and \( d \) is the actual distance between the slits.
04

Calculate Slit Separation (d)

For the first order bright fringe \( m = 1 \), substituting values, we have \[ 0.01 = \frac{1 \times 500 \times 10^{-9} \times 0.9}{d} \].Solve for \( d \): \[ d = \frac{500 \times 10^{-9} \times 0.9}{0.01} = 4.5 \times 10^{-5} \text{ m} = 45 \text{ micrometers}. \]
05

Use the Concept of Missing Orders

The third bright bands' absence indicates that the widths of the slits (a) cause them to coincide with minima of the diffraction pattern. The missing order condition is given by \[ a = \frac{m \lambda}{d} \], where \( m = 3 \) for the missing third fringe. Thus,\[ a = \frac{3 \times 500 \times 10^{-9}}{4.5 \times 10^{-5}} = 3.33 \times 10^{-5} \text{ m} = 33.3 \text{ micrometers}. \]

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Key Concepts

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

Understanding Wavelength
The concept of wavelength, especially in the context of light, is crucial to understanding various optical phenomena such as interference. Wavelength is typically defined as the distance between consecutive peaks or troughs in a wave. When dealing with light, we often measure this in nanometers (nm) due to its small size. In our exercise, we're working with a wavelength of 500.0 nm, which, when converted, is 500.0 x 10^-9 meters. This tiny measurement is a key factor in determining how light waves will interact with each other as they pass through the slits to form an interference pattern on a screen. Knowing the wavelength allows us to predict the behavior of light as it diffracts.
Diffraction Pattern
Diffraction occurs when light waves bend around obstacles or spread out after passing through narrow openings like slits. This spreading of light creates a pattern, known as a diffraction pattern, on a screen or surface. In a double-slit experiment, the diffraction pattern results from the combination of light waves emanating from each slit. These waves interfere with one another, leading to areas of constructive interference (bright fringes) where they overlap in phase, and areas of destructive interference (dark bands) where they are out of phase. The pattern's visibility helps us understand how light behaves in different conditions and is essential for determining the measurements like the slit width and separation.
Fringe Separation Explained
Fringe separation refers to the distance between consecutive bright or dark bands (fringes) on the screen. For a double-slit interference setup, this separation is determined by several factors: the light's wavelength, the distance between the slits, and the distance from the slits to the screen.The formula that describes fringe separation is:\[ y = \frac{m \lambda L}{d} \]Where:- \( y \) is the fringe separation- \( m \) represents the order number of the fringe- \( \lambda \) is the wavelength of the light- \( L \) is the distance to the screen- \( d \) is the separation between the slitsIn our problem, the first-order bright fringe separation \( y \) is given as 1.00 cm. This information allows us to solve for the slit separation, crucial for understanding the setup of the experiment.
Insight into the Missing Order Condition
In some interference patterns, you might notice that certain expected bright fringes are absent. This phenomenon is due to the missing order condition, which occurs when the slit width causes specific maxima from one slit to coincide with minima from the other, thereby canceling each other out.The missing order condition is mathematically expressed by the formula:\[ a = \frac{m \lambda}{d} \]Where:- \( a \) is the width of the slit- \( m \) is the order number, which in this case is 3, as the third bright bands are missing- \( \lambda \) is the wavelength- \( d \) is the separation between the slitsUnderstanding this condition is key to determining the slit's physical properties in an experiment, as observed when certain bright fringes do not appear on the pattern.

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Most popular questions from this chapter

If you can read the bottom row of your doctor's eye chart, your eye has a resolving power of 1 arcminute, equal to \(\frac{1}{60}\) degree. If this resolving power is diffraction limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume \(\lambda=550 \mathrm{nm} .\)

Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 km. If they broadcast \(3.6-\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

Light of wavelength 585 nm falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be? (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda=587.9782 \mathrm{nm}\) and \(\lambda=\) 587.8002 \(\mathrm{nm} .\) You wish to resolve these spectral lines in first order using a grating 1.20 \(\mathrm{cm}\) in length. What minimum number of slits per centimeter must the grating have?

A slit 0.240 \(\mathrm{mm}\) wide is illuminated by parallel light rays of wavelength 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 3.00 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(6.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

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