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Chapter 36: Problem 21

Mica Flake A thin flake of mica \((n=1.58)\) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe \((m=7)\) before the mica was used. If \(\lambda=550 \mathrm{~nm}\), what is the thickness of the mica? (Hint: Consider the wavelength of the light within the mica.)

Short Answer

Expert verified
The thickness of the mica is approximately 2436.7 nm.

Step by step solution

01

Determine the Wavelength in Mica

Calculate the wavelength of light within the mica using the formula \[ \lambda' = \frac{\lambda}{n} \] where \( \lambda = 550 \text{nm} \) and \( n = 1.58 \). \[ \lambda' = \frac{550 \text{ nm}}{1.58} \]
02

Calculate the Wavelength in Mica

Perform the division to find the wavelength of light within the mica. \( \lambda' = 348.1 \text{ nm} \)
03

Calculate Optical Path Difference

Determine the optical path difference created by the mica. Since the central point is now the seventh bright fringe, it means there is a path difference of 7 wavelengths in air. The formula for the optical path difference (OPD) is \[ \text{OPD} = 7 \lambda = 7 \times 550 \text{ nm} \]
04

Convert OPD to Thickness of Mica

Convert the optical path difference in air to the thickness of mica (t) using the relationship \[ t = \frac{\text{OPD}}{n} \] Where OPD is 7 times 550 nm, and n is 1.58. \[ t = \frac{7 \times 550 \text{ nm}}{1.58} \]
05

Calculate Thickness of Mica

Substitute the values into the formula to find the thickness of the mica. \[ t = \frac{3850 \text{ nm}}{1.58} \approx 2436.7 \text{ nm} \]

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Key Concepts

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

Optical Path Difference
In double-slit interference, optical path difference (OPD) plays a crucial role. OPD is the difference in travel paths of light waves due to a varying medium. When a thin layer like mica is introduced over one slit, it changes the effective distance that light travels. This difference shifts interference patterns.
Wavelength in Medium
To understand how wavelength changes in a medium, it is essential to know that light slows down in materials. Wavelength in a medium is given by lambda_prime: \(\backslash\frac{\backslash\text{lambda}}{\backslash\text{n}}\). When light moves from air to mica (refractive index 1.58), its wavelength decreases to \(348.1 \text{ nm}\). This new wavelength is vital for calculating optical phenomena.
Thickness Calculation
The thickness of mica is found by converting the optical path difference using the formula: \( \backslash t = \backslash\frac{\backslash\text{OPD}}{\backslash\text{n}} \). Given that the seventh bright fringe implies a path difference of seven wavelengths in air, we use this to compute the mica thickness. The result, 2436.7 nm, ensures precise calculation in practical applications.

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Most popular questions from this chapter

Use Phasors Add the quantities $$ \begin{array}{l} y_{1}=10 \sin \omega t \\ y_{2}=15 \sin \left(\omega t+30^{\circ}\right) \\ y_{3}=5.0 \sin \left(\omega t-45^{\circ}\right) \end{array} $$ using the phasor method.

Plane Wave A plane wave of monochromatic light is incident normally on a uniform thin film of oil that covers a glass plate. The wavelength of the source can be varied continuously. Fully destructive interference of the reflected light is observed for wavelengths of 500 and \(700 \mathrm{~nm}\) and for no wavelengths in between. If the index of refraction of the oil is \(1.30\) and that of the glass is \(1.50\), find the thickness of the oil film.

Long-Range Radio Waves Sources \(A\) and \(B\) emit long-range radio waves of wavelength \(400 \mathrm{~m}\), with the phase of the emission from \(A\) ahead of that from source \(B\) by \(90^{\circ} .\) The distance \(r_{A}\) from \(A\) to a detector is greater than the corresponding distance \(r_{B}\) by \(100 \mathrm{~m}\). What is the phase difference at the detector?

Two Glass Plates Two glass plates are held together at one end to form a wedge of air that acts as a thin film. A broad beam of light of wavelength \(480 \mathrm{~nm}\) is directed through the plates, perpendicular to the first plate. An observer intercepting light reflected from the plates sees on the plates an interference pattern that is due to the wedge of air. How much thicker is the wedge at the sixteenth bright fringe than it is at the sixth bright fringe, counting from where the plates touch?

Two Interference Patterns In a double-slit experiment the distance between slits is \(5.0 \mathrm{~nm}\) and the slits are \(1.0 \mathrm{~m}\) from the screen. Two interference patterns can be seen on the screen: one due to light with wavelength \(480 \mathrm{~nm}\), and the other due to light with wave length \(600 \mathrm{~nm}\). What is the separation on the screen between the third-order \((m=3)\) bright fringes of the two interference patterns?
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