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Nearly monochromatic coherent light waves leave two rectangular slits in phase and at an angle of \(\theta=22.0^{\circ}\) with the normal. When the light reaches a distant screen, the waves from the center of one slit are \(344^{\circ}\) out of phase with the waves from the center of the other slit, and the waves from the top of either slit are \(172^{\circ}\) out of phase with the waves from the bottom of that slit. (a) How is the center-to-center distance between the slits related to the width of either slit? (b) Calculate the intensity at the screen for \(\theta=22.0^{\circ}\) if the intensity at \(\theta=0^{\circ}\) is \(0.234 \mathrm{~W} / \mathrm{m}^{2}\)

Short Answer

Expert verified
The center-to-center distance between the slits is twice the width of either slit. The light intensity at \(\theta=22.0^{\circ}\) is approximately \(0.117 \mathrm{~W}/\mathrm{m}^{2}\).

Step by step solution

01

Understand the Relationship of the Center-to-center Distance and Width of a Slit

Monochromatic and coherent light waves when pass through rectangular slits create phase differences. When the light hits the distant screen, it's stated that waves from the centre of one slit are \(344^{\circ}\) out-of-phase with waves from the centre of the other slit. This means the path difference is equivalent to \(344^{\circ}\) i.e. \(344^{\circ} = (344/360)*\lambda\). But we know that path difference is calculated as \(d\sin{\theta}\) where \(d\) is the center-to-center distance between the slits, and \(\sin{\theta}\) is the angle of waves to the normal. Thereby comparing the two, \(d\sin{\theta} = (344/360)*\lambda\).
02

Calculate the Phase Difference from the Top and Bottom of Slit

It's mentioned that waves from the top or bottom of any slit are \(172^{\circ}\) out of phase with the waves from the center of that slit. Therefore, the path difference is equivalent to \(172^{\circ}\) i.e. \(172^{\circ} = (172/360)*\lambda\). However, the path difference also equals the slit width \(a\) times \(\sin{\theta}\), which leads to \(a\sin{\theta} = (172/360)*\lambda\).
03

Setting up the Relation

We now have the expressions \(d\sin{\theta} = (344/360)*\lambda\) and \(a\sin{\theta} = (172/360)*\lambda\). Dividing these two equations, we get \(d/a = 344/172 = 2\). So, the center-to-center distance between the slits is twice the width of either slit.
04

Calculate the Intensity at the Screen

The intensity \(I\) at \(\theta=22.0^{\circ}\) can be calculated with the formula \(I = I_0 cos^2(\Delta \phi / 2)\), where \(I_0\) is the maximum intensity at \(\theta=0^{\circ}\) and \(\Delta \phi\) is the phase difference. Given that \(I_0 = 0.234 \mathrm{~W} / \mathrm{m}^{2}\) and \(\Delta \phi = 344^{\circ} = (344 * \pi / 180)\), substitute these values into the formula to find the intensity at \(\theta=22.0^{\circ}\).
05

Calculate Intensity at a Given Angle

By substituting these values into the formula, the intensity \(I\) at \(\theta=22.0^{\circ}\) is \[I = 0.234 \mathrm{~W}/\mathrm{m}^{2} * cos^2((344 * \pi / 180)/2) = 0.234 \mathrm{~W}/\mathrm{m}^{2} * cos^2(3 \pi)\]. This leads us to a final intensity calculation of \(I ≈ 0.117 \mathrm{~W}/\mathrm{m}^{2}\).

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Key Concepts

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

Monochromatic Light
Monochromatic light refers to light that consists of a single wavelength or color. This is an important concept in various experiments involving light interference and diffraction. Light emitted from a monochromatic source does not change its wavelength. This is what makes it particularly useful in experiments like a double-slit experiment. For these exercises, we often use lasers because they generate monochromatic light.
Using monochromatic light ensures that the pattern observed on the screen is due to interference alone, without the complications of multiple wavelengths. In the context of interference of light, having a single wavelength makes it simpler to calculate and observe the resultant phase differences between waves. This uniformity is crucial in obtaining clear and predictable interference patterns. Understanding monochromatic light allows us to isolate and explore how light waves interact without the added variables introduced by multiple wavelengths.
Coherent Light
Coherent light is crucial for studying interference patterns in experiments. Coherent light waves maintain a constant phase relationship with one another over time. This consistency means that the light waves are in sync with each other.
In experiments like the double-slit experiment, coherence ensures the light waves will interact in a predictable manner. The result of this interaction can form areas of constructive interference, where the light waves amplify each other, and destructive interference, where they cancel each other out.
The sources of coherent light are usually laser beams. These beams are capable of producing the necessary constant phase differences required for interference patterns. In simple terms, coherent light can be thought of as rhythmic waves, each wave crest and trough perfectly aligned, leading to clear and consistent interference results.
Double-Slit Experiment
The double-slit experiment is a foundational experiment in understanding wave interference phenomena. It involves passing light through two close, narrow slits onto a screen.
As light passes through these slits, it diffracts and spreads out, causing the waves from each slit to overlap. This overlapping leads to an interference pattern on the screen that can be observed. Through this setup, we can observe bright and dark fringes. Bright fringes occur where constructive interference happens, where two waves add together to make a wave of greater amplitude, while dark fringes occur where destructive interference happens, where two waves cancel each other out.
The double-slit experiment beautifully demonstrates key principles of wave theory and light's wave-particle duality. Originally conducted by Young, this experiment vividly illustrates the interference of light, showing how wave properties play a role in how light behaves.
Phase Difference
Phase difference refers to the difference in the phase of two waves meeting at a point. In terms of light, the phase difference is crucial in determining whether two light waves interfere constructively or destructively.
In interference patterns created in experiments like the double-slit experiment, the phase difference between light coming from different slits determines the positions of bright and dark fringes on a screen. For example, if two waves are perfectly in phase (a phase difference of 0°), they will constructively interfere, resulting in a bright fringe. On the other hand, a phase difference of 180° will cause destructive interference, leading to a dark fringe.
Phase difference is often calculated considering the path difference the waves have traveled. When waves from two sources travel different distances to meet at a point on a screen, they may be out of phase if their path difference equates to a fractional wavelength. But if the path difference is a multiple of the wavelength, the two waves will be in phase, leading to constructive interference. This concept helps us calculate and predict the pattern of interference fringes in various optical setups.

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Most popular questions from this chapter

Monochromatic light of wavelength \(580 \mathrm{nm}\) passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at \(\pm 90.0^{\circ},\) so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta=45.0^{\circ}\) to the intensity at \(\theta=0 ?\)

Light of wavelength \(633 \mathrm{nm}\) from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide, and the resulting diffraction pattern is observed on a screen \(3.50 \mathrm{~m}\) away. What is the distance between the two dark fringes on either side of the central bright fringe?

A diffraction grating has 650 slits \(/ \mathrm{mm}\). What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm}\).)

The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about \(5.0 \mu \mathrm{m}\) in diameter) limits the size of an object at the near point \((25 \mathrm{~cm})\) of the eye to a height of about \(50 \mu \mathrm{m}\). (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye. eter of the human pupil is about \(2.0 \mathrm{~mm}\), does the Rayleigh criterion allow us to resolve a \(50-\mu \mathrm{m}\) -tall object at \(25 \mathrm{~cm}\) from the eye with light of wavelength \(550 \mathrm{nm} ?\) (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the \(25 \mathrm{~cm}\) near point with light of wavelength \(550 \mathrm{nm} ?\) (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes \(\left(60 \mathrm{~min}=1^{\circ}\right),\) and compare it with the experimental value of about \(1 \mathrm{min.}\) (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

Monochromatic light of wavelength \(\lambda=620 \mathrm{nm}\) from a distant source passes through a slit \(0.450 \mathrm{~mm}\) wide. The diffraction pattern is observed on a screen \(3.00 \mathrm{~m}\) from the slit. In terms of the intensity \(I_{0}\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maxi- mum: (a) \(1.00 \mathrm{~mm}\) (b) \(3.00 \mathrm{~mm}\) (c) \(5.00 \mathrm{~mm} ?\)

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