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3 years ago

5

You illuminate a slit with a width of 70.3 μm with a light of wavelength 719 nm and observe the resulting diffraction pattern on

a screen that is situated 2.11 m from the slit. What is the width, in centimeters, of the pattern?

1 answer:

Zielflug [23.3K]2 years ago

8 0

Answer:

4.3 cm

Explanation:

We are given that

Width,d=70.3 $\mu m=70.3\times10^{-6} m$

$1\mu m=10^{-6} m$

Wavelength, $\lambda=719 nm=719\times 10^{-9} m$

$1nm=10^{-9} m$

$r=2.11 m$

We have to find the width in cm of the pattern.

The angle for the first minimum m=1

$sin\theta_{min}\approx \theta=\frac{\lambda}{d}=\frac{719\times 10^{-9}}{70.3\times 10^{-6}}=0.0102 rad$

$y=r\theta=2.11\times 0.0102=0.0215 m$

The width of the pattern= $2y=2\times 0.0215=0.043 m=0.043\times 100=4.3 cm$

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Answer:

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Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

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u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

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