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kozerog [31]
3 years ago
15

Light of wavelength λ travels through a medium with an index of refraction n1before striking a thin film with an index of refrac

tion n2 at an angle of incidence of 0. Some of the light is reflected off of and some is transmitted through the n2 thin film. The transmitted light travels a distance t through the thin film before encountering the n1 medium again. Some of the light reflects off of the n1 medium.What is the effective path length difference between the light that reflected off of the n2 medium and the light that reflected off the n1 medium, given that n1>n2?
A. t+λ/n2
B. 2t+λ/n2
C. 2t+λ/(2n2)
D. t+λ/(2n2)
Physics
1 answer:
ra1l [238]3 years ago
5 0

Answer:

option C

Explanation:

The correct answer is option C

A light that transmits through n₂ travels t distance before reflection off the n₁ medium and again travels distance t before reaching the point from where it entered n₂  medium. Hence it travels 2 t distance more than the light that is reflected off n₂.

It( light entering n₂) also travels an additional distance equal to, half of the wavelength, when reflected off n₁ ( as n₁ is greater than n₂).  

Wavelength in n₂ is = \dfrac{\lambda}{n_2}  

Hence, path length difference = 2t +\dfrac{\lambda}{2 n_2}

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Makovka662 [10]

Answer:

45.3°C

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305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

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If you can throw a stone straight up to height h, what’s the maximum horizontal distance you could throw it over level ground?
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A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is
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Answer:

M is equal to m

Explanation:

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In any collision, as it is asumed that no external forces can act during the collision, momentum must be conserved.

So, if we call p₁ to the momentum before collision, and p₂ to momentum after it, taking into account the information above, we can write the following:

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From the question, we also know that it was an elastic collision.

In elastic collision, added to the momentum conservation, it must be conserved the kinetic energy also.

So, if we call k₁ to the kinetic energy prior the collision, and k₂ to the one after it, we can write the following:

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In this type of collision, it is said that the energy transfers from one mass to the other.

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Answer:

9s

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