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Sholpan [36]
2 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?
Physics
1 answer:
Vsevolod [243]2 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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Answer:

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Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

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Answer:

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