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

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by --------eqn 1

The angle through which the first minima is diffracted is given by ---------eqn 2

As is very small so we can write

So from eqn 1 and eqn 2 we can write

--------eqn 3

Here 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

Now putting all these value in eqn 3

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop