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Sindrei [870]
3 years ago
5

A very thin oil film (n = 1.25) floats on water (n = 1.33). What is the thinnest film that produces a strong reflection for gree

n light with a wavelength of 500 nm?
Physics
1 answer:
mixer [17]3 years ago
6 0

Answer:

200 nm is the thinnest film that produces a strong reflection for green light with a wavelength of 500 nm

Explanation:

If two reflected waves interfere constructively ,strong reflection is produced. Two reflected waves will experience a phase change

For constructive interference

2\times n\times t=m\lambda

for thinnest film m=1

refractive index should be taken for film n=1.25

thickness of the thinnest film is

t=\frac{m\lambda}{2n} \\t=\frac{1\times 500}{2\times 1.25} \\t=200 nm

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Answer:

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We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}, and y axle:0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}. Now replacing the value given as: v_{1i}=500(m/s), \beta_{1}=+60^{o} for the projectile proton and according to the problem \beta_{1}and\beta_{2} are perpendicular so \beta_{2}=-30^{o}, and assuming that m_{1}=m_{2}, we get for x axle:500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2} and y axle: 0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}, then solving for v_{2f}, we get:v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f} and replacing at the first equation we get:500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}, now solving for v_{1f}, we can find the speed of the projectile proton after the collision as:v_{1f}=250(m/s) and v_{2f}=\sqrt{3}*v_{1f}=433(m/s), that is the speed of the target proton after the collision.

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