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A glass with a refractive index of 1.50 is inserted into one arm of a Michelson interferometer that uses a 600.-nm light source. This causes the fringe pattern to shift by exactly 1000 fringes. How thick is the glass?

Short Answer

Expert verified
Answer: The thickness of the glass plate is 600 碌m.

Step by step solution

01

1. Write down the given values

We have been given the following information: - Refractive index of glass, n = 1.50 - Wavelength of light source, 位 = 600 nm - Fringe shift observed, 螖饾憮 = 1000 fringes
02

2. Formula for path difference

The path difference has a direct relationship with the fringe shift observed. For the given fringe shift, we can use the following formula for path difference: 螖d = 位 * 螖f
03

3. Calculate path difference

Now using the given values, let's calculate the path difference due to the glass plate: 螖d = (600 nm) * (1000 fringes) = 600,000 nm
04

4. Formula for thickness using refractive index

When a glass plate is inserted in the path, the light travels an additional distance through the glass plate. And the extra distance is related to the refractive index and thickness of the glass. The extra path difference can be given by: 螖饾憫 = 2 * (n-1) * t where t is the thickness of the glass plate.
05

5. Solve for thickness

Now, we can solve the above equation for the thickness (t) of the glass plate: t = 螖饾憫 / 2 * (n-1) t = (600,000 nm) / (2 * (1.50 - 1)) t = 600,000 nm / 1 = 600,000 nm
06

6. Convert the thickness to a suitable unit

Finally, let's convert the thickness to a more suitable unit such as micrometers: t = 600,000 nm * (1 碌m / 1,000 nm) = 600 碌m Therefore, the thickness of the glass plate inserted in the Michelson interferometer is 600 碌m.

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Key Concepts

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

Refractive Index
A refractive index is a number that describes how light propagates through a particular medium. When light enters a new medium, like glass, it changes speed, which results in bending or refracting the light. This is related to Snell's law, which describes how the angle of incidence is different from the angle of refraction due to the refractive index.
The refractive index (n) of a medium like glass is calculated by comparing the speed of light in a vacuum (approximately 3 imes 10^8 meters per second) to the speed of light in that medium.
  • If n > 1, light moves slower in the medium compared to a vacuum.
  • For the exercise, the refractive index of glass is given as 1.50.
This means light moves 1.5 times slower in glass than in a vacuum, causing light to bend and affect the path length, which is critical in devices like the Michelson interferometer.
Optical Path Difference
Optical path difference, often denoted as 螖d, is a measure of how much longer one travel path of light is compared to another in an interferometer. It's important in understanding interference patterns.
In the case of a Michelson interferometer, a beam of light is split into two paths that reflect back and recombine, creating an interference pattern. Any change in the length of these paths results in a shift in the interference fringes.
  • The exercise calculates the path difference as 螖d = 位 * 螖f, where 位 = 600 nm and 螖f = 1000 fringes.
  • Thus, 螖d = 600,000 nm gives us the extra distance traveled by light due to insertion of the glass plate.
Understanding the optical path difference is crucial for calculating how physical elements inserted into the path, like our glass plate, will affect the observed interference patterns.
Fringe Pattern Shift
Fringe pattern shift refers to the movement of interference fringes seen in devices like the Michelson interferometer. These fringes are dark and bright bands created by the constructive and destructive interference of light waves.
When a glass plate is introduced in the path of one of the beams, this affects the path length and refractive index, leading to a measurable shift in the fringe pattern.
  • The number of fringes moved indicates the change in path length caused by the glass plate.
  • In the exercise, a shift of 1000 fringes is directly linked to the optical path difference.
Calculating the fringe shift helps to accurately determine the thickness of materials, such as the glass plate in our exercise, based on the known values of light wavelength and refractive index. This practical application highlights the sensitivity and precision of the Michelson interferometer.

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Most popular questions from this chapter

Coherent monochromatic light passes through parallel slits and then onto a screen that is at a distance \(L=2.40 \mathrm{~m}\) from the slits. The narrow slits are a distance \(d=2.00 \cdot 10^{-5} \mathrm{~m}\) apart. If the minimum spacing between bright spots is \(y=6.00 \mathrm{~cm},\) find the wavelength of the light.

The thermal stability of a Michelson interferometer can be improved by submerging it in water. Consider an interferometer that is submerged in water, measuring light from a monochromatic source that is in air. If the movable mirror moves a distance of \(d=0.200 \mathrm{~mm},\) exactly \(N=800\) fringes move by the screen. What is the original wavelength (in air) of the monochromatic light?

For a double-slit experiment, two 1.50 -mm wide slits are separated by a distance of \(1.00 \mathrm{~mm} .\) The slits are illuminated by a laser beam with wavelength \(633 \mathrm{nm} .\) If a screen is placed \(5.00 \mathrm{~m}\) away from the slits, determine the separation of the bright fringes on the screen.

Suppose the distance between the slits in a double-slit experiment is \(2.00 \cdot 10^{-5} \mathrm{~m} .\) A beam of light with a wavelength of \(750 \mathrm{nm}\) is shone on the slits. What is the angular separation between the central maximum and the adjacent maximum? a) \(5.00 \cdot 10^{-2} \mathrm{rad}\) b) \(4.50 \cdot 10^{-2} \mathrm{rad}\) c) \(3.75 \cdot 10^{-2} \mathrm{rad}\) d) \(2.50 \cdot 10^{-2} \mathrm{rad}\)

Two different wavelengths of light are incident on a diffraction grating. One wavelength is \(600 . \mathrm{nm}\) and the other is unknown. If the 3 rd order of the unknown wavelength appears at the same position as the 2 nd order of the \(600 . \mathrm{nm}\) light, what is the value of the unknown wavelength?

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