The thickness is 155 nm at t1.
The thickness is 77.3 nm at t2.
The inquiry informs us that the laser light's wavelength is λ=510nm
The plastic rod's refractive index is n=1.30
The transparent coating's refractive index is nr=1.65
Minimum reflection would be required for maximal light transmission into the rod, and it is mathematically described as
2t1=510+10⁻⁹/1.65
t1=510+10⁻⁹/1.65*2
t1=155nm
where m is the interference order, which equals 1.
2t2= {m+1/2} λ/nr
The thickness is t replacing values
t1=155 nm
The highest reflection would occur for minimal light penetration through the rod, and this maximum reflection is mathematically described as
2t2= [m+1/2] λ/nr
t2=77.3 nm
The complete question is- Laser light of wavelength 510 nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has n = 1.30. The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?
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2. Is right..... if not then go with 1
<u>Answer:</u>
<em>The average speed of the car is 66.9 km/h</em>
<u>Explanation:</u>
Here distance covered with the speed <em>57 km/h=7 km </em>
distance covered with the speed of <em>81 km/h=7 km</em>
<em>Average speed is equal to the ratio of total distance to the total time.
</em>
<em>total distance= 7 + 7= 14 km </em>
<em>
</em>
<em>time taken to cover the first 7 km= 7/57 h </em>
<em>time taken to cover the second part of the journey = 7/81 h
</em>
<em>average speed = 
</em>
<u><em>Shortcut:
</em></u>
<em>When equal distances are covered with different speeds average speed=2 ab/(a+b) where a and b are the variable speeds in the phases.
</em>
Answer:
If the boat is going 36 km/h, then it will take about 5.8 hours to travel a distance of 210 km/h.
Explanation:
To calculate time, you have to divide the distance by the speed.
In this case, we take 210 and divide it by 36. The answer you will get is 5.83...
Answer:
v = 21 m / s
Explanation:
We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data
v = v₀ + a t
the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s
a = (v-v₀) / t
a = (11 - 6) / 8
a = 0.625 m / s²
when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s
v = v₀ + a t
v = 11 + 0.625 16
v = 21 m / s
