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AfilCa [17]
3 years ago
15

A glass lens, nglass = 1.52, has a 140 nm thick antireflective film coating one side, nfilm = 1.37. White light, moving through

the air, is perpendicularly incident on the coated side of the lens. What is the largest wavelength of the reflected light that is totally removed by the coating?
Physics
1 answer:
Paul [167]3 years ago
5 0

Answer:

767.2 nm

Explanation:

The optical path for the reflection for the inner surface is:

Path=2\times n_{film}\times thickness

For maximum wavelength, this distance has to be traveled doubled (The ray is normally incident). So,

\lambda=4\times n_{film}\times thickness

Given that:

n_{film}=1.37

Thickness = 140 nm

So, wavelength is:

\lambda=4\times 1.37\times 140\ nm

<u>Wavelength is 767.2 nm</u>

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3 years ago
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Consider two different rods. The greatest thermal conductivity will be in the rod with:
sergey [27]

Answer:

Options A and D are correct

Explanation:

The thermal conductivity of a metal is the property of a metal to allow heat flow through it. conductivity is higher in conductors and low in insulators. Thermal conductivity is high in metals due to the metallic bonds that exist in metals and the presence of free electrons within the metal which allow easy flow of heat from one atom to another.From the problem the rod which contains freer electrons will allow more heat to flow easily hence have a higher thermal conductivity.

Thermal conductivity has the formula below;

k= \frac{QL}{AΔT}

  • k is thermal conductivity,
  • A is cross sectional area
  • L is length of rod
  • Q is quantity of heat transferred to material.
  • ΔT is temperature change.

From the above equation we can see that thermal conductivity is inversely proportional to A and directly proportional to L. This mean the rod with less area will have a higher thermal conductivity and the rod with a higher length will have higher k. Hence option C i wrong and option D is correct.

For specific heat, its very much different from thermal conductivity. Specific heat is the ability of a material to hold heat while thermal conductivity is the ability of heat to flow through a material.

4 0
3 years ago
Give the SI unit. For physics
IrinaVladis [17]

Answer:

metre (m) - unit of length

kilograms (kg) - unit of mass

second (s) - unit of time

ampere (A) - unit of electrical current

kelvin (K) - unit of temperature

mole (mol) - unit of the amount of substance

Explanation:

6 0
3 years ago
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jeka94

Answer:

What does that even mean?

Explanation:

7 0
3 years ago
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An airplane has a mass of 1.9E6 kg and the air flow past the
nordsb [41]

Answer:

v_{1}=164.4 m/s  

Explanation:

The Bernoulli equation is:

P+\frac{1}{2}\rho v^{2}+\rho gh=constant (1)

  • P is pressure related to the fluid
  • ρ is the density of the fluid (ρ(air)=1.23 kg/m³)
  • v is the speed of the fluid
  • h is the displacement from one position to the other

Now let's apply the equation (1), for our case:

P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2} (2)

here we assume that h is the same in both cases so it canceled out.

  • <u>subscript 1</u> is related to the upper surface of the wings
  • <u>subscript 2</u> is related to the low surface of the wings

Solving the equation for v₁ we have:

v_{1}=\sqrt{\frac{2(P_{2}-P_{1})}{\rho}+v_{2}^{2}} (3)

Now, we know that pressure P=F/A (force over area)

\Dela P=\frac{W}{A}=\frac{mg}{A}=\frac{1.9\cdot 10^{6}\cdot 9.81}{1.6\cdot 10^{3}} =11649.4 N/m^{2} (4)

Combining (3) and (4), we can find v1.

v_{1}=\sqrt{\frac{2(11649.4)}{1.23}+90^{2}}  

v_{1}=164.4 m/s  

I hope it helps you!  

6 0
3 years ago
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