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saveliy_v [14]
4 years ago
9

A very thin film of soap, of thickness 170 nm, in between air seems dark. On the other hand, when placed on top of glass some vi

sible light is seen to shine from the film. How can this happen and what is the smallest visible light that creates constructive interference when we place the film on top of glass
Physics
1 answer:
torisob [31]4 years ago
5 0

Answer:

λ₀ = 2 d n

Explanation:

A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.

Let's analyze the general conditions of this interference,

* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º

* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.

          λ = λ₀ / n

where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains

         2d = (m + ½) λ

the expression for destructive interference remains

         2d = m λ

          2d = m λ₀ / n

When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.

Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.

Find us the wavelength that this constructive interference creates

           2d n = m λ₀

           λ₀ = 2 d n / m

To find the minimum wavelength, suppose we observe the first interference pattern m = 1

         λ₀ = 2 d n

where d is the thickness of the film and n the index of refraction of the same

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Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

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                                  x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2

                                 F =ma

                                 V(t) = V_{o} +at

                                 x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2

(a) Calculating velocity right before going up the ramp.

 Wooden block is going on a straightaway and has net for on it.

         F_{n} =F-F_{s} = F-uF_{n}  = 100N-0.4*9.8m/s^2*5kg =81N

     and this force produces acceleration of

      a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .

With this acceleration, wooden block would reach at the foot of ramp in.

          x(t) = 12m = 16.2m/s^2*t^2

         t = 1.217s

and final velocity will be

v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

                              V= V_{h}cos(20) = 18.5288m/s

and deceleration along the ramp is

                              a = \frac{F_{s} }{m}

 Where F_{s} force of friction along the inclined plane.

F_s =  uF_n = u*m*a

a = 9.8m/s^2*cos(20) = 9.2089m/s^2

is a component of g along normal of the inclined plane.

                               F_{s} = 0.25*5kg*9.2089m/s^2

                              = 11.5112N

                              a = \frac{11.5112N}{5kg} = 2.3022m/s^2

And with this deceleration time needed to get wooded block to stop is.

                     v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0

                        t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s

 and in that time wooden block would travel

   x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m

This is how up wooden box will go before coming to stop.

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