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Ymorist
15 days ago
5

Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizon

tally, and close enough to each other so that both of them fall in the beam. A precise positioning shield is used to cover alternatively one of the slits, while the other remains open. On an observation screen positioned 1 meter further from the opaque screen, the diffraction patterns from the independent slit illumination were found identical, with minima 4.0 mm apart. The interference pattern maxima from simultaneous illumination of both slits (shield removed) were 1.0 mm apart. What are the double slit characteristics (slit width, slit separation)? Provide the result in millimeters.
i found that the answers for part 1: slit width= 0.158 mm slit separation= 0.633 mm

problem 2: On the setup described in problem 1, the opaque screen with slits was replaced with a diffraction grating with 20 slits/mm. What distance between the diffraction maxima is to be expected on the observation screen?
Physics
1 answer:
Softa [2K]15 days ago
6 0

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

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Solution:

Em_{f} / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

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14 days ago
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Answer:

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