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Light from a helium-neon laser \((\lambda=632,8 \mathrm{~nm})\) is incident on a single slit. What is the maximum width of the slit for which no diffraction minima are observed?

Short Answer

Expert verified
The maximum width of the slit for which no diffraction minima are observed is approximately \(632.8 \times 10^{-9}\) m.

Step by step solution

01

Identify the given quantities

The wavelength of the laser light (\(\lambda\)) is given as 632.8 nm. The task is to find the maximum width of the single slit (\(d\)) for which no diffraction minima are observed.
02

Understand the condition for a diffraction minimum

A diffraction minimum occurs when the path difference between waves from different parts of the slit is an integer multiple of the wavelength. This is when \(d \sin(\theta) = m\lambda\), where \(m\) is an integer not equal to 0. If no diffraction minima are to be observed, then for the smallest possible minimum (where \(m=1\)), \(\sin(\theta)\) must be greater than 1 which is not possible. Hence the condition for no diffraction minima is that \(d < \lambda\). Hence the maximum value of \(d\) is just less than \(\lambda\).
03

Convert wavelength from nm to m

As the SI unit of length is meter(m), we should convert the wavelength from nm to m in order to provide the answer in the right units. 1nm = \(1 \times 10^{-9}\) m so \(\lambda = 632.8 \times 10^{-9}\) m.
04

Calculate the slit width

As explained in step 2, the width of the slit (\(d\)) without diffraction minima should be less than the wavelength (\(\lambda\)). Therefore, the maximum width of the slit is just less than \(\lambda\), so approximately \(d \approx 632.8 \times 10^{-9}\) m.

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Key Concepts

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

Single Slit Diffraction
When light encounters an obstacle or passes through a narrow aperture, such as a single slit, it bends around the edges and spreads out. This phenomenon is known as diffraction. In the case of single slit diffraction, the light waves passing through the slit interfere with one another creating a pattern of bright and dark fringes on a screen placed behind the slit. The central bright fringe is the most intense and is flanked by alternate dark and light fringes.

The dark fringes, known as minima, occur at specific angles where the light waves from different parts of the slit destructively interfere. The condition for these minima can be described by the equation:
\( d \sin(\theta) = m\lambda \)
where \(d\) is the slit width, \(\theta\) is the angle of incidence, \(\lambda\) is the wavelength of light, and \(m\) is the order of the minima (an integer). Understanding this principle is crucial for analyzing the behavior of light in various applications, from scientific instruments to optical technologies.
Wavelength of Laser Light
The wavelength of laser light is a critical parameter in evaluating diffraction patterns. Laser light is chosen for experiments like these because it's highly monochromatic, meaning the light consists of essentially one wavelength. The wavelength determines the spacing of the diffraction pattern's fringes.

For a helium-neon laser, commonly used in educational settings, the wavelength is typically about 632.8 nm (nanometers), where 1 nm is equivalent to \(1 \times 10^{-9}\) meters. This particular wavelength falls within the red part of the visible light spectrum. The precise knowledge of the wavelength allows for the calculation of diffraction patterns and helps in understanding how light will behave when encountering objects of comparable sizes to the wavelength.
Condition for No Diffraction Minima
The textbook exercise focuses on the condition for observing no diffraction minima. This condition arises when the width of the slit is narrower than the wavelength of the incident light.

To avoid diffraction minima, the path difference between the waves emerging from different parts of the slit must be less than one wavelength so that they cannot destructively interfere to create a minimum. The equation providing this condition is \(d < \lambda\), where \(d\) is the slit width and \(\lambda\) is the wavelength of light. Consequently, when the slit width is smaller than the wavelength, the light does not have sufficient space to spread out and interfere, thus eliminating the occurrence of dark fringes or minima in the diffraction pattern.
Path Difference in Waves
The concept of path difference plays a pivotal role in the interference and diffraction of light. Path difference refers to the difference in distance traveled by two waves arriving at the same point. In the context of single slit diffraction, waves emanating from different parts of the slit travel different distances to reach the same point on a screen.

If this path difference is equal to an integer multiple of the wavelength (\(m\lambda\)), the waves will destructively interfere and produce a minimum. For constructive interference and bright fringes (maxima), the path difference must be an integer multiple of the wavelength plus half its value (\((m + \frac{1}{2})\lambda\)). Understanding path difference is essential to grasp how diffraction patterns form and how to manipulate them for various scientific and technological purposes.

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Most popular questions from this chapter

Intense white light is incident on a diffraction grating that has 600 lines/mm. (a) What is the highest order in which the complete visible spectrum can be seen with this grating? (b) What is the angular separation between the violet edge \((400 \mathrm{~nm})\) and the red edge \((700 \mathrm{~nm})\) of the first-order spectrum produced by the grating?

The critical angle for total internal reflection for sapphire surrounded by air is \(34.4^{\circ} .\) Calculate the Brewster's angle for sapphire if the light is incident from the air.

\(\ln\) a location where the speed of sound is \(354 \mathrm{~m} / \mathrm{s}\), a \(2000-\mathrm{Hz}\) sound wave impinges on two slits \(30.0 \mathrm{~cm}\) apart. (a) At what angle is the first maximum located? (b) If the sound wave is replaced by \(3.00-\mathrm{cm}\) microwaves, what slit separation gives the same angle for the first maximum? (c) If the slit separation is \(1.00 \mu \mathrm{m}\), what frequency of light gives the same first maximum angle?

In a Young's interference experiment, the two slits are separated by \(0.150 \mathrm{~mm}\) and the incident light includes two wavelengths: \(\lambda_{1}=540 \mathrm{~nm}\) (green) and \(\lambda_{2}=\) \(450 \mathrm{~nm}\) (blue). The overlapping interference patterns are observed on a screen \(1.40 \mathrm{~m}\) from the slits. (a) Find a relationship between the orders \(m_{1}\) and \(m_{2}\) that determines where a bright fringe of the green light coincides with a bright fringe of the blue light. (The order \(m_{1}\) is associated with \(\lambda_{1}\), and \(m_{2}\) is associated with \(\lambda_{2} .\) (b) Find the minimum values of \(m_{1}\) and \(m_{2}\) such that the overlapping of the bright fringes will occur and find the position of the overlap on the screen.

Waves from a radio station have a wavelength of \(900 \mathrm{~m}\). They travel by two paths to a home receiver \(20.0 \mathrm{~km}\) from the transmitter. One path is a direct path, and the second is by reflection from a mountain directly behind the home receiver. What is the minimum distance from the mountain to the receiver that produces destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)

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