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elena55 [62]
1 year ago
9

monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central

maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm. find the wavelength of the light.
Physics
1 answer:
hjlf1 year ago
5 0

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit  is 0.75 so  Wavelength of the light  is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity =  1.35 mm

λ =  wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2  

=5.0625 *10^-7 = 506.25

so

Wavelength of the light 506.25.

To learn more about Wavelength of the light refer to:

brainly.com/question/15413360

#SPJ4

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A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. What total distance does the m
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Explanation:

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A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

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angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

                           = T/2 = 0.952/2 = 0.476 kips/in²

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Answer: the answer is d

Explanation:

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