Question 1

In a thin film experiment, a wedge of air is used between two glass plates.If the wavelength of the incident light in air is 480 nm, how much thicker is the air wedge at the 16th dark fringe than it is at the 6th?

Accepted Answer

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Question 2

An air wedge is formed using two glass plates that are in contact along their left edge.When viewed by highly monochromatic light, there are exactly 4001 dark bands in the reflected light.The air is now evacuated (with the glass plates remaining rigidly fixed)and the number of dark bands decreases to exactly 4000.The index of refraction of the air is:

Accepted Answer

The change in the number of dark bands is due to the change in the optical path length caused by the removal of air. The index of refraction of air is slightly greater than 1, and the decrease in dark bands from 4001 to 4000 indicates that the index of refraction of air is approximately 1.00025.

Question 3

An air wedge is formed from two glass plates which are in contact at their left edges.There are ten dark bands when viewed by reflection using monochromatic light.The left edge of the top plate is now slowly lifted until the plates are parallel.During this process:

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Question 4

Yellow light is viewed by reflection from a thin vertical soap film.Let$\lambda$ be the wavelength of the light within the film.Why is there a large dark space at the top of the film?

Accepted Answer

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Question 5

In a Michelson interferometer, in order to shift the pattern by half a fringe, one of the mirrors at the end of an arm must be moved by:

Accepted Answer

In a Michelson interferometer, the phase difference between the two beams is given by:

$\Delta\varphi = \frac{4\pi}{\lambda}(x_2-x_1)$

where $\lambda$ is the wavelength of the light, $x_2$ is the distance traveled by the light in the arm with the movable mirror, and $x_1$ is the distance traveled by the light in the arm with the fixed mirror.

When the movable mirror is moved by a distance $\Delta x$, the phase difference changes by:

$\Delta\varphi = \frac{4\pi}{\lambda}(\Delta x)$

To shift the pattern by half a fringe, we need to change the phase difference by $\frac{\pi}{\lambda}$. Therefore:

$\frac{\pi}{\lambda} = \frac{4\pi}{\lambda}(\Delta x)$

$\Delta x = \frac{1}{4}\lambda$

This means that to shift the pattern by half a fringe, we need to move the mirror by a distance equal to one-quarter of the wavelength. Therefore, the correct answer is (C).

Question 6

A thin film with an index of refraction of 1.60 is placed in one of the beams of a Michelson interferometer.If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 580 nm, what is the thickness of the film?

Accepted Answer

The shift in the fringe pattern is given by:
$\Delta m = \frac{2t}{\lambda}(n-1)$ where $t$ is the thickness of the film, $\lambda$ is the wavelength of light, $n$ is the refractive index of the film and $\Delta m$ is the number of fringes shifted.

Rearranging the equation, we get:
$t = \frac{\Delta m \lambda}{2(n-1)}$

Substituting the given values, we get:
$t = \frac{8\cdot 580	imes 10^{-9}}{2(1.6-1)} = 3.9\ \mu m$

Therefore, the thickness of the film is 3.9 µm, which is option C.

Question 7

A liquid of refractive index n = 4/3 replaces the air between a fixed wedge formed from two glass plates as shown.As a result, the spacing between adjacent dark bands in the interference pattern:

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Quiz 35: Interference 1

In a thin film experiment, a wedge of air is used between two glass plates.If the wavelength of the incident light in air is 480 nm, how much thicker is the air wedge at the 16th dark fringe than it is at the 6th?

An air wedge is formed from two glass plates which are in contact at their left edges.There are ten dark bands when viewed by reflection using monochromatic light.The left edge of the top plate is now slowly lifted until the plates are parallel.During this process:

Yellow light is viewed by reflection from a thin vertical soap film.Let λ\lambdaλ be the wavelength of the light within the film.Why is there a large dark space at the top of the film?

In a Michelson interferometer, in order to shift the pattern by half a fringe, one of the mirrors at the end of an arm must be moved by:

A thin film with an index of refraction of 1.60 is placed in one of the beams of a Michelson interferometer.If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 580 nm, what is the thickness of the film?

A liquid of refractive index n = 4/3 replaces the air between a fixed wedge formed from two glass plates as shown.As a result, the spacing between adjacent dark bands in the interference pattern: