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3 years ago

9

Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a

s each needle is less than a millimeter thick. So, to find the thickness, she uses a needle to poke a hole in a piece of brown construction paper. Then, she positions a 640 nm laser pointer to shine through the hole and project a circular diffraction pattern on a wall 21.7 m away. She then uses her ruler to measure that the central bright circle is 14.2 cm in diameter. What diameter does Irina calculate for the needle?

1 answer:

Vsevolod [243]3 years ago

8 0

Answer:

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by $d sin\Theta =1.22\lambda$ --------eqn 1

The angle through which the first minima is diffracted is given by $tan\Theta =\frac{y_1}{D}$ ---------eqn 2

As $\Theta$ is very small so we can write $sin\Theta =tan\Theta$

So from eqn 1 and eqn 2 we can write

$y_1=\frac{1.22\lambda D}{d}$ --------eqn 3

Here $y_1$ is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so $y_1=\frac{14.7}{2}=7.35 \ cm$

Now putting all these value in eqn 3

$d=\frac{1.22\lambda D}{y_1}$

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

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A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

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q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

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where

m is the mass of the particle

v is its final speed

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$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

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So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

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