8 points)Antireflection coating can be used on the eyeglasses to reduce the reflection of light: a) A 100nm thick coating is applied to the lens. What must be the coating’s index of refraction to be most effective at 500nm? (Assume the coating index of refraction is less than that of the lens). b) If the index of refraction of the coating is 1.20, find the necessary thickness of the coating at 500nm.

Answers

Answer 1

Answer:

- the coating’s index of refraction is 1.25

- the required thickness is 104.1667 nm

Explanation:

Given the data in the question;

Thickness of coating t = 100 nm

wavelength λ = 500nm

we know that refractive index is;

t = λ/4n

make n, the subject of formula

t4n = λ

n = λ / 4t

we substitute

n = 500 / ( 4 × 100 )

n = 500 / 400

n = 1.25

Therefore, the coating’s index of refraction is 1.25

2)

given that;

Index of refraction of the coating; n = 1.20

λ = 500 nm

thickness of coating t = ?

t = λ / 4n

we substitute

t = 500 / ( 4 × 1.2 )

t = 500 / 4.8

t = 104.1667 nm

Therefore, the required thickness is 104.1667 nm


Related Questions

Verify that your equation has the masses and the velocities before and after the collision. Solve equation for the initial velocity of the projectile, Vo. As the bob swings upward from h1 to a max of h2, what is happening to the kinetic energy of the system?

Answers

Answer:

Decrease occur in kinetic energy.

Explanation:

The kinetic energy decreases when the the Bob swings reaches to the maximum height because the motion of the bob slower down. At maximum height, the kinetic energy decreases whereas the value of potential energy is the highest. The main reason of higher potential energy is that it depends on height of an object while on the other hand, kinetic energy depends on the motion of an object so that's why the value of kinetic energy decreases and potential energy increases at maximum height of the bob.

A system can only achieve a lower energy state by:_______

Answers

Transferring its energy to its surroundings.

An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the point of origin of the flight. The plane flies with an airspeed of 120 m/s. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination

Answers

Answer:

The right solution is "4.8° east of north".

Explanation:

Given:

Distance,

= 500 km

Speed,

[tex]\vec{v}=120 \ m/s[/tex]

Wind (towards west),

[tex]v_0=10 \ m/s[/tex]

According to the question, we get

The angle will be:

⇒ [tex]\Theta=Cos^{-1}(\frac{v_0}{v_1} )[/tex]

       [tex]=Cos^{-1}(\frac{10}{120} )[/tex]

       [tex]=85.21[/tex] (north of east)

hence,

The direction must be:

⇒ [tex]\Theta'=90-85.21[/tex]

        [tex]=4.79^{\circ}[/tex]

or,

        [tex]=4.8^{\circ}[/tex] (east of north)

How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?

Answers

Answer:

(a) v₁ = 51.96 km/h

(b) v₁ = 178 km/h

Explanation:

(a)

For having the same momentum:

m₁v₁ = m₂v₂

where,

m₁ = mass of Volkswagen = 816 kg

v₁ = speed of Volkswagen = ?

m₂ = mass of Cadillac = 2650 kg

v₂ = speed of Cadillac = 16 km/h

Therefore, using these values in the equation, we get:

[tex](816\ kg)v_1 = (2650\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{2650\ kg}{816\ kg})[/tex]

v₁ = 51.96 km/h

(b)

For having the same momentum:

m₁v₁ = m₂v₂

where,

m₁ = mass of Volkswagen = 816 kg

v₁ = speed of Volkswagen = ?

m₂ = mass of Truck = 9080 kg

v₂ = speed of Truck = 16 km/h

Therefore, using these values in the equation, we get:

[tex](816\ kg)v_1 = (9080\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{9080\ kg}{816\ kg})[/tex]

v₁ = 178 km/h

A man standing on a frictionless ice throws a 1.00kg mass at 20m/s at an angle elevation of 40.0 degrees. What was the magnitude of the mans momentum immediately after the the throl

Answers

Answer:

Explanation:

1.00kg×20m/s×cos40=15.3

calculate the electric potential 3mm from a point charge of 16Nc​

Answers

[tex]4.8 \times 10^8[/tex] volts

Explanation:

The electric potential due to a point charge is given by

[tex]V= \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q}{r}[/tex]

where Q = charge = [tex]16 \times 10^{-9}[/tex] C

r = distance from a point = [tex]3 \times 10^{-3}[/tex] m

[tex]\varepsilon_{0}[/tex] = permitivity of free space

= 8.85×10^-12 C^2/N-m^2

Plugging in the numbers,

[tex]V = \dfrac{1}{4 \pi (8.85 \times 10^{-12})} \dfrac{16 \times 10{-9}}{3 \times 10^{-3}}[/tex]

[tex]= 4.8 \times 10^8[/tex] volts

A 16 kg science book is dropped of a 120 meter high cliff. Assuming a closed system:
a) how fast is book traveling the instant before it impacts the ground below the cliff?
b) how far above the bottom of the cliff is the object moving at 12 m/s?

Answers

Answer:

Explanation:

The mass of that science book...wow. In pounds that would be 35.2! Yikes!

Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is

[tex]v^2=v_0^2+2a[/tex]Δx  where

v is the final velocity, our unknown,

v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,

a is the pull of gravity which is always -9.8 m/s/s, and

Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:

[tex]v^2=0^2+2(-9.8)(-120)[/tex] so

[tex]v=\sqrt{2(-9.8)(-120)}[/tex] and

v = 48 m/s

As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:

[tex]12^2=0^2+2(-9.8)[/tex]Δx and

144 = -19.6Δx so

Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:

120 - 7.3 = 112.7 m off the ground.

Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second.

Answers

Answer:

[tex]F = 14.2 N[/tex]  

Explanation:

From the question we are told that:

Mass [tex]m=0.15kg[/tex]

Radius [tex]r=0.6[/tex]

Angular Velocity [tex]\omega=2rev/s[/tex]

                            [tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]

Generally the equation for Force applied is mathematically given by

 [tex]F =mrw2[/tex]

 [tex]F=0.15*0.6* (4*x3.14^)2[/tex]

 [tex]F = 14.2 N[/tex]    

A horizontal force is applied to a 4.0 kg box. The box starts from rest, moves a horizontal distance of 10.0 meters, and obtains a velocity of 7.0 m/s. The change in the kinetic energy is:_____.

Answers

Answer:

98 J

Explanation:

Applying,

Change in kinetic energy = Final kinetic energy- initial kinetic energy

ΔK.E = mv²/2-mu²/2..............Equation 1

Where ΔK.E = Change in kinetic energy, m = mass of the box, u = initial velocity of the box, v = final velocity of the box.

From the question,

Given: m = 4.0 kg, u = 0 m/s, v = 7 ,0 m/s

Substitute these values into equation 1

ΔK.E = (4(7²)/2)-(4(0²)/2)

ΔK.E = (2×49)-0

ΔK.E = 98 J

Hence the change in kinetic energy 98 J

A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.

Answers

Answer:

The speed of projection is 34 m/s.

Explanation:

Height of building, h = 51 m

horizontal distance, d = 74 m

time, t = 8 s

Let the angle is A and the speed is u.

d = u cos A x t

74 = u cos A x 8

u cos A = 9.25 .... (1)

Use second equation of motion

[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]

Squaring and adding both the equations

[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]

a car increases its speed as it moves across the floor. which form of energy is increasing for the car?

Answers

Answer:

kinetic

Explanation:

i just remember it from last year

Answer:

kinetic energy

Explanation:

expression for kinetic energy is

kinetic energy = (1/2) × mass × (velocity)^2

so , as velocity increases K.E increases

We say that evaporation is a cooling process. (a) What cools and what warms? We say that condensation is a warming process. (b) What warms and what cools?

Answers

Answer:

a liquid warms gas cools

b solid piece warms and liquid cools

Kinematics equations tells us the position of an object under constant acceleration increases linearly with time.
A. True
B. False

Answers

Answer:

False.

Explanation:

Suppose that we have an object that moves with constant acceleration A.

Then the acceleration of the object is defined by the equation:

a(t) = A

The acceleration is the rate of change of the velocity, then the velocity equation is given by the integration of the acceleration equation, we will get:

v(t) = A*t + V₀

Where V₀ is the velocity of the object at the time t = 0s.

Now, if we integrate it again, we will get the position equation:

p(t) = (1/2)*A*t^2 + V₀*t + P₀

Where P₀ is the initial position equation.

Here, we can see that the position equation is a quadratic equation (not a linear equation), then the statement is false.

explain how a lever can act as a force multiplier

Answers

Answer:

Example:Opening of a bottle cap by tool

when we hold a tool and open the bottle cap this is because , force x tool force .

The load arm is usually shorter than the effort arm in second order levers. Moving a large weight hence requires less effort. A force multiplier lever or effort multiplier lever is the name for this kind of lever. A boat's oars, for instance, can increase the force.

What is second order levers?

Second-order levers are devices with the input force farthest from the fulcrum and the output force on the same side of the fulcrum. A wheelbarrow is an excellent illustration of a second-order lever.

A second-order lever will have an output force greater than an input force, similar to first-order levers. The output journey, however, will be shorter than the input length. Both the input and output forces in this situation will move in the same direction.

Learn more about lever here:

https://brainly.com/question/18937757

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At a playground, Maryam a 3-year old girl and Zahirah a 6-year old girl are playing with the swings. Maryam is sitting while Zahirah is standing on the swing. Both of them were given the same push by their mother. Choose the CORRECT statements:

A. Maryam is swinging faster than Zahirah.
B. Zahirah is swinging faster than Maryam.
C. Both swings at the same pace.
D. Maryam is swinging faster since she is younger.
E. Zahirah is swing faster since she is older.​

Answers

Answer:

both swings at the same place

Explanation:

because there mother is giving same amount of force to both.

E. Is the answer. Explaination: I have a 3 year old and I would not push her as high or as fast as a six year old.

A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.

Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?

Answers

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

Newton's Second Law of Motion
14. Why was it necessary to transfer mass from the cart to the hanger rather than just
simply adding mass to the hanger?

Answers

Answer:

The weight of the hanging mass provides tension in the string, which helps to accelerate the cart along the track. A small frictional force will resist this motion. 

.Use Newton's third law to describe the forces that are exerted by the falling egg and the ground. Explain how the use of the straws in the design affects the forces

Answers

Answer:

Newton's third law states that for every action, there is an equal and opposite reaction. That means when you exert a force on an object, the object exerts a force back on you. ... Using shock-absorbing materials can help reduce the amount of force exchanged between the ground and the egg.

Explanation:

I hope it helps you

When one of the tall straws hits the ground, the energy is transferred to the center of the pyramid and then to the egg, but as the middle straw is connected to the outer surface of the egg, energy enters trying to make the egg rotate. This is governed by Newton's third law.

What is Newton's third law?

If an object exerts a force on another object, then another object must exert a force of equal magnitude and opposite direction back on first object.

What are examples of Newton's third law?

Examples of Newton's third law are:

A swimmer moves forward by pushing off the side of pool. This way, the wall pushes in opposite direction and giving acceleration.

Another example is rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the rocket. This reaction force is called thrust.

Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and forward motion.

To learn more about Newton's third law here

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The sound from a trumpet radiates uniformly in all directions in 20C air. At a distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB. The frequency is 587 Hz. (a) What is the pressure amplitude at this distance

Answers

Answer:

The answer is below

Explanation:

The intensity level (B) of a sound wave is given by:

B = 10log(I/I₀);

where I₀ is the threshold intensity = 1 * 10⁻¹² W/m², I is the intensity at distance 5 m, B is the intensity level = 52 dB

Substituting gives:

[tex]52=10log(\frac{I}{10^{-12}} )\\\\log(\frac{I}{10^{-12}} )=5.2\\\\I=1.58*10^{-7}\ W/m^2[/tex]

The pressure is given by:

[tex]I=\frac{p_{max}^2}{2\rho v} \\\\\rho=air\ density=1.2\ kg/m^3,v=speed\ of\ sound\ in\ air=344\ m/s,p_{max}=pressure:\\\\p_{max}=\sqrt{2\rho vI}=\sqrt{2*1.58*10^{-7}*1.2*344} =1.14*10^{-2}Pa[/tex]

Why did vygotsky believe that children were more dependent learners

Answers

Answer:

Vygotsky's sociocultural theory asserts that learning is an essentially social process in which the support of parents, caregivers, peers and the wider society and culture plays a crucial role in the development of higher psychological functions.

Thus,  Vygotsky believed that children were more dependent learners

A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?

Answers

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

A temperature of 200 degrees Fahrenheit is equivalent to approximately A.93.3 degrees Celsius B. 232 degrees Celsius C. 37.8 degrees Celsius D. 840 degrees Celsius

Answers

Answer:

you can use G.oogle for this question.

Answer:

93.3 degrees Celsius.

Explanation:

Give a quantitative definition of being in contact.

Answers

Two things are said to be in contact if the smallest distance between a point in one of them and a point in the other one is zero.

It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?

Answers

Earth has its own atmosphere. That is one reason all the water that has been on Earth has been recycled through the water cycle. It never leaves Earth’s atmosphere.

6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if its surface normal isin the XY plane and along a line that isinclined at 60 degrees to the positive Y axisand 30 degrees to the positive X axis

Answers

Answer:

Flux is 21 Nm^2/C.

Explanation:

Electric field, E = 6 N/C along X axis

Electric filed vector, E = 6 i N/C

Area, A = 4 square meter

Area vector

[tex]\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\[/tex]

The flux is given by

[tex]\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} . \left (3.5 \widehat{i} + 2 \widehat{j} \right )\\\\\phi = 21 Nm^2/C[/tex]

here did you walk? What did you find most enjoyable while walking: listening to music, listening to an audio book, or nothing? How did your body react to this introductory amount of exercise? Was it more exercise or less exercise than you are used to? If you did not walk, what other type of physical movement did you do?

Answers

Working out in the guy

A electron gains electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?

a. V1 = V2
b. V1 > V2
c. V1 < V2

Answers

Answer:

We know that the change in electric potential energy is defined as:

q*ΔV = ΔP

So, the change in the electric potential energy is the charge times the change in the electric potential.

For the case of an electron gas, we have:

q = -e

where -e is the charge of an electron (remember that is negative).

So, if the electron gains electric potential then:

ΔP > 0

this means that the final potential energy is larger than the initial one, then we have:

-e*ΔV > 0

This means that ΔV must be negative.

V₂ = electric potential at point 2, so it is the final electric potential

V₁ = electric potential at point 1, so it is the final electric potential

Then we should get:

ΔV = V₂ - V₁ < 0.

This means that:

V₂ < V₁

The correct option is b.

How does the density of water change when: (a) it is heated from 0o
C to
4o
C; (b) it is heated from 4o
C to 10o
C ?

Answers

Answer:

[b] it id heated from 4o

Explanation:

On a day when the speed of sound is 345 m/s, a 440 Hz tuning fork causes a tube closed at one end to vibrate in the second harmonic. How long is the tube?​

Answers

Answer:

Length = 3.136 meters

Explanation:

Given the following data;

Speed = 345 m/s

Frequency = 440 Hz

To find how long is the tube;

First of all, we would determine the wavelength;

Wavelength = speed/frequency

Wavelength = 345/440

Wavelength = 0.784 m

Next, we would determine how long is the tube using the formula;

[tex] Length = \frac {2n - 1}{4} * wavelength [/tex]

Substituting into the formula, we have;

[tex] Length = \frac {2*1 - 1}{4} * 0.784 [/tex]

[tex] Length = \frac {2 - 1}{4} * 0.784 [/tex]

[tex] Length = \frac {1}{4} * 0.784 [/tex]

Length = 3.136 meters

A 2.2 kg, 20-cm-diameter turntable rotates at 80 rpm on frictionless bearings. Two 600 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?

Answers

Answer:

[tex]w_2=38.3rpm[/tex]

Explanation:

From the question we are told that:

Mass of turntable [tex]M=2.2kg[/tex]

Diameter of turntable [tex]d=20cm=>0.2m[/tex]

Angular Velocity [tex]\omega =80rpm[/tex]

Mass of Blocks [tex]M_b=600g=>0.6kg[/tex]

Generally the equation for inertia is mathematically given by

Initial scenario at \omega=80rpm

 [tex]I_1=\frac{1}{2}mR^2[/tex]

 [tex]I_1=\frac{1}{2}*2.2*0.1^2[/tex]

 [tex]I_1=0.11kgm^2[/tex]

Final scenario

 [tex]I_2=I_1+2mR^2[/tex]

 [tex]I_2=0.011+(2*0.6*0.12)[/tex]

 [tex]I_2=0.023[/tex]

Generally the equation for The relationship between Angular velocity and inertia is mathematically given by

 [tex]I_1w_1=I_2w_2[/tex]

 [tex]w_2=\frac{I_1 \omega}{I_2}[/tex]

 [tex]w_2=\frac{0.011*80}{0.023}[/tex]

 [tex]w_2=38.3rpm[/tex]

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