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Solnce55 [7]
3 years ago
5

A Michelson interferometer operating at a 600 nm wavelength has a 3.59-cm-long glass cell in one arm. To begin, the air is pumpe

d out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028.How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Physics
1 answer:
Finger [1]3 years ago
7 0

Answer:

The number of bright dark fringe shifts is 35

Solution:

As per the question:

Wavelength of the light, \lambda = 600\ nm

Refractive index of air, \mu_{a} = 1.0028

Length, d = 3.59 cm

Now,

The no. of bright-dark fringe shifts can be calculated by the given formula:

m = (\mu - 1)\frac{2d}{\lambda}

Thus

m = (\mu_{a} - 1)\frac{2d}{\lambda}

m = (1.00028 - 1)\frac{2\times 3.59\times 10^{- 2}}{600\times 10^{- 9}}

m = 33.5 ≈ 34

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The normal force is the supporting force that is exerted on an object that is in contact with another stable object.

Answer: Option C

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3 years ago
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How does increasing energy affect the amplitude of a wave?​
Crazy boy [7]

Answer:

The amount of energy carried by a wave is related to the amplitude of the wave

Explanation:

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Hope this helped!!!

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3 years ago
When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where i
Gnoma [55]

Answer:

The ball stops instantaneously at the topmost point of the motion.

Explanation:

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The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.

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4 years ago
The texture of obsidian is best described as glassy.Why does obsidian contain few or no mineral grains? by the way I put it unde
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3 years ago
You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica
Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A} , where:

l is the length

m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3

We divide the first equation by the second equation to get:

\frac{m}{R} = \frac{d A^2}{\rho}

A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8}   \ m^2

Using this Area, we find the diameter of the wire:

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D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}

D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm

To find the length, we multiply the two equations stated initially:

mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}

l^2 = 8.534\\l =   2.92 \ m

8 0
3 years ago
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