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disa [49]
2 years ago
6

When light is reflected from a thin film of oil on a water puddle, the colors we see are produced by interference. What two wave

s are interfering in this situation? Explain.
Physics
1 answer:
Paraphin [41]2 years ago
6 0

The two waves are interfering with the reflections from the top and bottom of the oil film.

<h3>Thin Film Interference:</h3>

A natural phenomenon known as "thin-film interference" occurs when light waves reflected by a thin film's upper and lower limits collide with one another, either boosting or weakening the light that is reflected.

Interference is what gives an oil slick floating on water or a soap bubble in the light of their vibrant colors. Colors that interfere positively are the ones with the brightest hues. Thin film interference is the name given to the phenomenon because it occurs when light reflected from many thin film surfaces interferes with one another.

Because light waves reflecting from the top surface of a film interfere with waves reflecting off the bottom surface, this phenomenon is known as thin-film interference.

Learn more about thin film interference here:

brainly.com/question/17074217

#SPJ4

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A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops
Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).  

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg andv_{2} =0.4v_{1} from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the T_{sat}=179.88^{0} C \leq T_{1} from the table 1 we get:v_{1} =0.30661(m^{3}/Kg). Now we have second conditions as: P_{2}=1(MPa), T_{2}=250^{0}C so from the same table we can see the state still superheated and we getv_{2}=0.23275(m^{3}/Kg), knowing that it's a isobaric process we can find the compression's work as:W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ) so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and:W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ) and the second process is isochoric:W_{2}=zero,nowW_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ). Finally to get the temperarure at the final state in part (b) we get:v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa) from table 2 (see attached) we comparev_{f} andv_{g} at the saturated water table and find the following:v_{f}=0.001093(m^{3}/Kg), so we know that the final state phase is a satured mixture and we get the temperature at the final state as:T_{2} =T_{sat} =151.83^{0}C.

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