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nignag [31]
3 years ago
6

A light beam travels at 1.94×108 in quartz. The wavelength of the light in quartz is 355 .Part AWhat is the index of refraction

of quartz at this wavelength?Part BIf this same light travels through air, what is its wavelength there?
Physics
1 answer:
Alja [10]3 years ago
4 0

A) 1.55

The speed of light in a medium is given by:

v=\frac{c}{n}

where

c=3\cdot 10^8 m/s is the speed of light in a vacuum

n is the refractive index of the material

In this problem, the speed of light in quartz is

v=1.94\cdot 10^8 m/s

So we can re-arrange the previous formula to find n, the index of refraction of quartz:

n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55

B) 550.3 nm

The relationship between the wavelength of the light in air and in quartz is

\lambda=\frac{\lambda_0}{n}

where

\lambda is the wavelenght in quartz

\lambda_0 is the wavelength in air

n is the refractive index

For the light in this problem, we have

\lambda=355 nm\\n=1.55

Therefore, we can re-arrange the equation to find \lambda_0, the wavelength in air:

\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm

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Conclusions are showed as follows:

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Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
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Solution:

As per the question:

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Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

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