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horrorfan [7]
4 years ago
12

A Michelson interferometer uses light with a wavelength of 452.446 nm. Mirror M2 is slowly moved a distance x, causing exactly 2

.2700×102 bright-dark-bright fringe shifts to be observed starting and ending with a central bright spot. If the mirror moved only half the distance, what would the final image be?
a. Bright spot.

b. Dark spot.

c. In between bright and dark spots.

d. Impossible to determine, more information needed.
Physics
1 answer:
I am Lyosha [343]4 years ago
3 0

Answer:

Option b. Dark spot

Explanation:

Michelson interferometer gives the path difference of the light as path difference, d as twice the distance moved or covered by the mirror M_{2}, x and is given as:

d= 2x

Since, its an even multiple, we obtain a bright fringe

now, when the move half the distance, i.e., \frac{x}{2}

Therefore, the path difference for half the distance \frac{x}{2} is:

d = x

As it is clear that its an odd multiple which correspond to dark spot as the final image

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Answer:

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Explanation:

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3 years ago
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.12 m whose potential is 230
slamgirl [31]

Answer:

(a) q=3.07 nC

(b) σ=17 nC/m²

Explanation:

Given data

Radius r=0.12m

Potential V=230 V

To find

(a) Charge q

(b) Charge density σ

Solution

For Part (a)

As we know that potential is:

V_{potential}=k_{constat}\frac{q_{charge}}{r_{radius}} \\\\q_{charge}=\frac{V_{potential*r_{radius}}}{k_{constant}}

Substitute the given values

q_{charge}=\frac{(0.12m)(230V)}{9*10^{9} }\\ q_{charge}=3.07*10^{-9}C\\or\\q_{charge}=3.07nC

For Part (b)

The charge density is given by:

σ=q/(4πr²)

Substitute the given values and value of q to find charge density

So

=\frac{3.07*10^{-9}C}{4\pi (0.12m)^2}\\ =1.69*10^{-8}C/m^2\\or\\=17nC/m^2

σ=17 nC/m²

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3 years ago
A light bulb that is connected to a 120 V source drawers 1.7 A. What is the resistance of the light bulb?
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