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gavmur [86]
3 years ago
13

For this problem, you will only be concerned with the geometric aspects of thin-film interference, so ignore phase shifts caused

by reflection from a medium with higher index of refraction. (Because of the structure of a butterfly's wings, such phase shifts do not contribute much to what you actually see when you look at the butterfly.) Part A Assume that light is incident normal to the surface of a film of thickness d. How much farther does the light reflected from the back surface travel than the light reflected from the front surface
Physics
1 answer:
Nana76 [90]3 years ago
4 0

Answer:

Λ = 2 t

Explanation:

This is a problem is interference by reflection, where it is requested not to take into account the changes of phase and change in the wavelength within the film

                   λₙ = λ₀ / n

When the wave enters the film, it must reach the end of the film, be reflected and reach the initial surface, therefore the total length is

                d = 2 t

where t is the thickness of the film, because the film is perpendicular to the surface

therefore the difference in optical path between the reflected ray on the surface and the end film is

                  Λ = 2 t

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VikaD [51]

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

<h3>What is  the standard normal curve?</h3>

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<h3>Use of the standard normal curve:</h3>

Use the normal distribution's standard form to calculate probability. Since the standard normal distribution is indeed a probability distribution, the probability that a variable will take on a range of values is indicated by area of the curve between two points. 100% or 1 is the total area beneath the curve.

<h3>According to the given data:</h3>

the region to the left of the standard normal curve,

z=0.49

To the right of,

z = 2.05

So,

The area will be:

= P[z < 0.49] + P[ z >2.05]

= P[z < 0.49] + 1 -  P[ z < 2.05]

= .6879 + 1 - .9798

= 0.7081

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

To know more about standard normal curve visit:

brainly.com/question/12972781

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I understand that the question you are looking for is:

Find the area under the standard normal curve to the left of z = 0.49 and to the right of z = 2.05. Round your answer to four decimal places, if necessary.

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If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later
garik1379 [7]
X=1/2 at^2
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If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!

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