1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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}

You might be interested in
The atomic nucleus contains two subatomic particles, the proton and the neutron. Atoms of different elements have different numb
9966 [12]
The answer is option A .

7 0
3 years ago
Read 2 more answers
A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ
siniylev [52]

Answer:

m=1.53kg    

Explanation:

To solve this problem we use the Hooke's Law:

F=k*\Delta x     (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm  

The force applied is due to the weight

F=mg

We replace in (1):

mg=k*\Delta x  

We solve the equation for m:

m=k*\Delta x/g=5*3/9.81=1.53kg    

7 0
3 years ago
A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo
gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

                       H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

 Height at any time 'T' after the drop =

                          (initial height) +

                                              (initial velocity) x (T) +
                                                                 (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

     Upward  = the positive direction

                       Initial height = +150 m
                       Initial velocity = + 3 m/s

     Downward = the negative direction

                     Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

                 H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
0    =  (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

                   -4.9 T²  +  3T  + 150  =  0

Use the quadratic equation:

                         T  =  (-1/9.8) [  -3 plus or minus √(9 + 2940)  ]

                             =  (-1/9.8) [  -3  plus or minus  54.305  ]

                             =  (-1/9.8) [ 51.305  or  -57.305 ]

                          T  =  -5.235 seconds    or    5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

                    H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
H'(t)  =  v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set                                  v₀ + a T  =  0

                                      +3  -  9.8 T  =  0

Add 9.8 to each  side:   3               =  9.8 T

Divide each side by  9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh !  I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
                       (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) =  1.5 m/s .

Distance climbed during that time = (average speed) x (time)

                                                           =  (1.5 m/s) x (0.306 sec)

                                                           =  0.459 meter  (hardly any at all)

     But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at  150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely  5.847 seconds after its release.  

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work.  The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0
3 years ago
Hi please, I Have An attachment on Waves, Just two Objective Questions Whoever Answers Will be Marked Brainliest thank you.
kifflom [539]

Answer:

The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.

5 0
3 years ago
A flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω. It is connected to a 21. 2-v battery at the insta
MissTica

For a  flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω, the rate of energy being delivered is mathematically given as

P= 53 W

<h3>What rate is energy being delivered by the battery?</h3>

Generally, the equation for the Battery power  is mathematically given as

P = I (dt)V

Therefore

P= 2.50 A * 21.2V

P= 53 W

In conclusion, rate of energy being delivered

P= 53 W

Read more about Energy

brainly.com/question/13439286

7 0
2 years ago
Other questions:
  • Machines
    7·2 answers
  • A loop of wire is placed in a uniform magnetic field. For what orientation of the loop is the magnetic flux a maximum? For what
    11·1 answer
  • . Draw the sketches of two waves A and B such that wave A has twice the wavelength and half the amplitude of wave B.
    15·1 answer
  • Urgent! Based on the diagram below, what color will each pigmented paper appear to be to an observer?
    5·1 answer
  • What is your hypothesis (or hypotheses) for this experiment?
    14·2 answers
  • What is the net force on this object?​
    5·1 answer
  • The movement of crustal plates is best described as a:
    10·2 answers
  • Calculate the force of an object that has a mass of 10kg and an acceleration of 4m/s²
    15·1 answer
  • MULTIPLE CHOICE
    15·1 answer
  • Define and Explain the Sovereignty
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!