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Chapter 3: Problem 76

Two microscope slides made of glass are illuminated by monochromatic \((\lambda=589 \mathrm{nm})\) light incident perpendicularly. The top slide touches the bottom slide at one end and rests on a thin copper wire at the other end, forming a wedge of air. The diameter of the copper wire is \(29.45 \mu \mathrm{m} .\) How many bright fringes are seen across these slides?

Short Answer

Expert verified
The number of bright fringes seen across the slides is approximately 100.

Step by step solution

01

Understanding the problem

We have two glass slides with air trapped between them, forming a wedge-shape. Since the air film's thickness varies between the slides, we will observe bright and dark fringes due to constructive and destructive interference.
02

Formulate the equation for constructive interference

For bright fringes to occur due to constructive interference, the condition is given by: \(2t=n\lambda\), where \(t\) is the thickness of the air gap, \(n\) is the integer representing the fringe position, and \(\lambda\) is the wavelength of light.
03

Express the thickness in terms of the wedge angle

The wedge angle, denoted by \(A\), is defined as the angle between the two slides. Considering the wedge as a right triangle, we have: \(t = x \tan A\), where \(x\) is the distance along the slides. From step 2, we have: \(2x \tan A = n\lambda\)
04

Calculate the wedge angle

We can find the wedge angle using the given diameter of the copper wire, \(d = 29.45\,\mu m\), and the length of the slides, \(L\), assuming the slides are much longer than their width. The wedge angle A can be found using the formula: \(A = \tan^{-1}(\frac{d}{L})\) As \(A\) is very small, we can approximate \(\tan A \approx A\). We have: \(A = \frac{d}{L}\)
05

Calculate the number of bright fringes

Substituting the expression of \(A\) from step 4 in the expression obtained in step 3, we have: \(2x \frac{d}{L} = n\lambda\) We need to find out the number of bright fringes, i.e., the value of \(n\) when \(x = L\). So, the expression becomes: \(2d = n\lambda\) Solving for \(n\), we have: \(n = \frac{2d}{\lambda}\) Now, we'll plug in the given values: \(n = \frac{2(29.45 \times 10^{-6} m)}{589 \times 10^{-9} m}\) Calculating this expression gives us: \(n \approx 100\) Therefore, the number of bright fringes seen across the slides is approximately 100.

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Most popular questions from this chapter

A hydrogen gas discharge lamp emits visible light at four wavelengths, \(\lambda=410,434,486,\) and \(656 \mathrm{nm}\) (a) If light from this lamp falls on a \(N\) slits separated by \(0.025 \mathrm{mm},\) how far from the central maximum are the third maxima when viewed on a screen \(2.0 \mathrm{m}\) from the slits? (b) By what distance are the second and third maxima separated for \(l=486 \mathrm{nm}\) ?

A thin film with \(n=1.32\) is surrounded by air. What is the minimum thickness of this film such that the reflection of normally incident light with \(\lambda=500 \mathrm{nm}\) is minimized?

(a) As a soap bubble thins it becomes dark, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the bubble can be and appear dark at all visible wavelengths? Assume the same index of refraction as water. (b) Discuss the fragility of the film considering the thickness found.

Shown below is a double slit located a distance \(x\) from a screen, with the distance from the center of the screen given by \(y .\) When the distance \(d\) between the slits is relatively large, numerous bright spots appear, called fringes. Show that, for small angles (where \(\sin \theta \approx \theta,\) with \(\theta\) in radians), the distance between fringes is given by \(\Delta y=x \lambda / d\)

Is there a phase change in the light reflected from either surface of a contact lens floating on a person's tear layer? The index of refraction of the lens is about \(1.5,\) and its top surface is dry.
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