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Sedbober [7]
3 years ago
14

Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin fi

lm of oil with n2 =1.75. On the other side of the thin film is glass with n3 =1.5. What is the minimum non-zero value of the film thickness d that will cause the reflections from both sides of the film to interfere constructively?
Physics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

Explanation:

On both sides of the film , the mediums have lower refractive index.

for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is

2μt = ( 2n +1 ) λ / 2

μ is refractive index of film ,t is thickness of film λ is wavelength of light

n is order of fringe

for minimum thickness

n = 0

2μt =  λ / 2

t =  λ / 4μ

= 670 / 1.75 x 4

= 95.71 nm .

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Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate s
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The POSITION of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the VELOCITY from A to B is/are the same as expressed in both coordinate systems.

Explanation:

Imagine that you got 2 coordinate systems; A and B. Now, this coordinate systems start at different points.

Measured from A, the starting point of the A coordinate system its, of course, at (0,0,0)_A. But B its somewhere else. Lets put the origin of the system B MEASURED from the system A, at (b_x,b_y,b_z)_A.

This position, in the system B will be at (0,0,0)_B, cause its it origin, be careful, we could write:

(b_x,b_y,b_z)_A = (0,0,0)_B

as this two vectors are the same, the starting position for the B system.

Lets imagine that in thei got something displaced from the starting point of the system B by (1,0,0). Of course, in the system B this is located at (1,0,0)_B, What is the position for the system A? Well, it will be at

(b_x,b_y,b_z)_A + (1,0,0)_A = (b_x +1,b_y,b_z)_A.

The distance is measured the same way in both system.

Any position

(x,y,z)_B

in the system B, will be at

(b_x,b_y,b_z)_A + (x,y,z)_A = (b_x + x,b_y + y,b_z + z)_A

at system A, cause this positions ARE THE SAME, even though they are represented with different vector in this systems.

Now, why the velocity its the same? Well, the velocity its the derivative of the position with respect time.

So, for a position R=(x,y,z)_B, in the system B we got:

\frac{d}{dt} R = \frac{d}{dt} (x,y,z)_B = ( \frac{dx}{dt} , \frac{dy}{dt} , \frac{dz}{dt} )_B

R in the system A takes the form:

R=(b_x,b_y,b_z)_A + (x,y,z)_A

and the derivative its:

\frac{d}{dt} R = \frac{d}{dt} [(b_x,b_y,b_z)+ (x,y,z)]_A =\frac{d}{dt} (b_x,b_y,b_z)_A + \frac{d}{dt}  (x,y,z)_A

now, the (b_x,b_y,b_z)_A its constant, so:

\frac{d}{dt} (b_x,b_y,b_z)_A  = 0

which leave us with:

\frac{d}{dt} R =\frac{d}{dt}  (x,y,z)_A

\frac{d}{dt} R = (\frac{d}{dt} x,\frac{d}{dt} y,\frac{d}{dt} z)_A

This is the same for both systems!

4 0
3 years ago
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