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allsm [11]
3 years ago
9

A thin flake of mica ( n = 1.58 ) is used to cover one slit ofa double slit interference arrangement. The central point on thevi

ewing screen is now occupied by what had been the seventh brightside fringe ( m = 7 ). If λ = 550 nm, what is thethickness of the mica?
Physics
1 answer:
ELEN [110]3 years ago
5 0

Answer:

the thickness of the mica is 6.64μm

Explanation:

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))

Where

L = Thickness

n = Index of refraction of each material

\lambda = Wavelength

Our values are given as

\Phi = 7(2\pi)L=tn_1 = 1.58n_2 = 1\lambda = 550nm

Replacing our values at the previous equation we have

\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))

t = \frac{7*550}{1.58-1}\\t = 6637.931nm \approx 6.64\mu m

the thickness of the mica is 6.64μm

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A loop of current-carrying wire has a magnetic dipole moment of 5. 0 10–4 am2. if the dipole moment makes an angle of 57° with a
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The potential energy will be 1.46*10^-4J.

To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.

<h3>How to find the potential energy of the loop?</h3>
  • We have the expression for torque acting on a current loop in a uniform magnetic field as,

                         \tau=MBsin\theta

where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.

  • As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
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            \tau=W=U=MBsin\theta=5*10^{-4}*0.35*sin57=1.46*10^{-4}J

Thus, we can conclude that, the potential energy will be 1.46*10^-4J.

Learn more about the torque here:

brainly.com/question/27949876

#SPJ4

7 0
1 year ago
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