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kobusy [5.1K]
3 years ago
9

A Michelson interferometer operating at a 500 nm wavelength has a 3.73-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?
Engineering
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

The number of bright-dark fringe is 42

Solution:

As per the question:

Wavelength of light, \lambda = 500\ nm = 500\times 10^{- 9}\ m

Length of the glass cell, x = 3.73 cm = 0.0373 m

Refractive index, \mu = 1.00028

Now,

To calculate the bright-dark fringe shifts, we use the formula given below:

d_{m} = \frac{2x}{\lambda }\times (\mu - 1)

Now, substituting the appropriate values in the above formula:

d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)

d_{m} = 41.77 ≈ 42

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Answer:

Correct option a) True.

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It is true since the Vickers hardness value refers to the force applied in a 136 ° diamond tip penetrator divided by the surface of the groove produced in the material, the lower the impression made on this greater the value will be end of the Vickers measurement and greater its hardness.

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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
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Answer:

a) \dot m_{3} = 135\,\frac{lbm}{s}, b) h_{3}=168.965\,\frac{BTU}{lbm}, c) T = 200.829\,^{\textdegree}F

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\dot m_{1} + \dot m_{2} - \dot m_{3} = 0

The mass flow rate exiting the tank is:

\dot m_{3} = \dot m_{1} + \dot m_{2}

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b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

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h_{1}=180.16\,\frac{BTU}{lbm}

h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)

h_{2}=29.032\,\frac{BTU}{lbm}

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h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }

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