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Trava [24]
3 years ago
5

Consider the interference/diffraction pattern from a double-slit arrangement of slit separation d = 6.60 um and slit width a. Th

e wavelength of the monochromatic light incident normally upon the slits is 2n = d/10 (in air). There is a filter (of negligible thickness) placed on slit 2, so that the magnitude of the EM wave emitted from it is half of that emitted from slit 1. The space between the slits and the screen is filled with water, whose index of refraction is n = 1.33 (you can take noir as 1.00).
(a) What is the wavelength 2 of the light in water?
(b) If a << 1, What is the phase difference between the waves from slits 1 and 2?
(c) For a << 2, Derive an expression for the intensity / as a function of O and other relevant parameters, including the intensity at the center of the screen (where 0 = 0).
(d) Now suppose a = d/3. Redo part (b) above. How many interference maxima are present within the central diffraction peak? (Do not count the "clipped" maxima, if any.) (4) — E -2 d E ) в /Б/ = 1/5 | TT

Physics
1 answer:
Digiron [165]3 years ago
3 0

Answer:

(a) λ = 0.496 um (b) S =2π Δ d sinθ/ λ  (c) I =gI₀ (d) For the central diffraction peak, a total of 5 interference maxima are present or available.

Note: find an attached copy of a part of the solution to the given question below.

Explanation:

Solution

Recall that:

d = 6.6 um

λ₀ =d/10

λ₀ = 6.6 um

Now,

(a) We find the wavelength λ of the light in water.

Thus,

λ water = (λ₀ )/n

= 0.66/1.33

So,

λ water = λ = 0.496 um

(b) We find the phase difference between the waves from slit 1 and 2

Now,

if a <<d  and a<<λ

Then the path difference between the rays will be

Δ S₂N = Δ d sinθ

Thus, the phase difference becomes,

S = 2π Δ/λ is S= 2π Δ d sinθ/ λ

<u>(</u>c) The next step is to derive an expression for the intensity  I as function of O and other relevant parameters.

Now,

Let p be the point where these two rays interfere with each other.

Thus,

The electric field vector coming out from slot and and slot 2 is

E₁= E₀₁ cos (ks₁ p - wt) i

E₂ = E₀₂ cos (ks₂ p - wt) i

Note: Kindly find an attached copy of a part of the solution to the given question below.

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(C) Average Velocity = 25 m/s due North

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A block of mass m is pushed a horizontal distance D from position A to position B, along a horizontal plane with friction coeffi
Wewaii [24]

Answer:

The total work done by friction is -2 · μ · m · g · D

Explanation:

Hi there!

The work done by a force is calculated as follows:

W = F · d · cos θ

Where:

W = work.

F = force that does the work.

d = displacement.

θ = angle between the displacement and the force.

If the force is horizontal, as in this case, cos θ = 1

The friction force is calculated as follows:

Ffr = μ · m · g

Where:

μ = friction coefficient.

m = mass of the object.

g = acceleration due to gravity.

Then, in this case, the work done by friction when pushing the block from A to B will be:

W AB = -Ffr · D

W AB = - μ · m · g · D

Notice that the friction force is negative because it is opposite to the pushing force P.

When the block is pushed from B to A, the work done by friction will be:

W BA = Ffr · (-D)

W BA = -μ · m · g · D

Now, the displacement is negative and the friction force is positive (in the opposite direction to -P).

The total work done by friction will be:

W AB + W BA = - μ · m · g · D  - μ · m · g · D  = -2 μ · m · g · D

5 0
3 years ago
Section 2.2
Feliz [49]

Answer:

Speed changes at the rate of 24 m/s for each second over time.

Explanation:

We are told the object's acceleration is equal to 24 m/s²

Now we know that acceleration can also be defined as the rate of change of speed with time. Also speed has a unit known as m/s.

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A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 12 A, while that
andreev551 [17]

Answer:

a) 1450watts

b) 564watts

c) 1.11

Explanation:

Power consumed = IV

I is the current rating

V is the operating voltage

If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;

a) For blow-dryer

Operating voltage = 120V

Its current rating = 12A

Power consumed = IV

= 120×12

= 1440watts

b) For vacuum cleaner:

Operating voltage is the same as that of blow dryer = 120V

Its current rating = 4.7A

Power consumed = IV

= 120×4.7

= 564watts

c) Energy used = Power consumed × time taken

Energy used = Power × time

Energy used by blow dryer = 1440×20×60

= 1,728,000Joules

Energy used up by vacuum cleaner = 564×46×60

= 564×2760

= 1,556,640Joules

Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11

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ratelena [41]

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