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Alja [10]
3 years ago
9

Light rays in a material with index of refraction 1.31 can undergo total internal reflection when they strike the interface with

another material at a critical angle of incidence. Find this material\'s index of refraction when the required critical angle is 78.3°.
Physics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

The refractive index of the material is 1.28.

Explanation:

It is given that,

Refractive index of medium 1, n₁ = 1.31

Critical angle, \theta_c=78.3

At critical angle rays will reflects at 90 degrees. Using Snell's law as:

n_1sin\theta_1=n_2sin\theta_2

At critical angle, n_1sin\ \theta_c=n_2\ sin90

n_1sin\ \theta_c=n_2

n_2=1.31\times sin(78.3)

n_2=1.28

So, the refractive index of the material is 1.28. Hence, this is the required solution.

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Answer: The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another. Another way of stating this law of chemistry is to say the total energy of an isolated system.

Explanation:

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In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the
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Answer:

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8 0
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6. State whether each of the following is an example of conduction, convection, and/or radiation. Explain if you think there is
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3 0
3 years ago
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
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