A parallel beam of light from a He-Ne laser, with a wavelength 633 nm, falls on two very narrow slits 0.070 mm apart.
Part A
How far apart are the fringes in the center of the pattern on a screen 4.1 m away?

Answers

Answer 1

The distance between the fringes in the center of the pattern on a screen 4.1 m away is approximately 0.032 mm.

The distance between the two slits, d, is given as 0.070 mm. The distance between the slits and the screen, L, is 4.1 m. The wavelength of the laser light, λ, is 633 nm.

The distance between the central maximum and the first-order maximum can be calculated using the formula:

y = (λL) / d

where y is the distance between the fringes.

Substituting the given values, we get:

y = (633 x 10^-9 m) x (4.1 m) / (0.070 x 10^-3 m)

y = 0.037 mm

This gives the distance between the central maximum and the first-order maximum. Since there is a fringe at the center, we need to subtract the distance between the two adjacent fringes to get the distance between the fringes in the center.

The distance between two adjacent fringes can be calculated as:

Δy = λL / d

Substituting the values, we get:

Δy = (633 x 10^-9 m) x (4.1 m) / (0.070 x 10^-3 m)

Δy = 0.005 mm

Therefore, the distance between the fringes in the center of the pattern is:

y - Δy = 0.037 mm - 0.005 mm

y - Δy = 0.032 mm

The distance between the fringes in the center of the pattern on a screen 4.1 m away is approximately 0.032 mm. The interference pattern is a result of the wave nature of light and the phenomenon of interference, where the light waves from the two slits interfere constructively and destructively to form a pattern of bright and dark fringes on the screen. The distance between the fringes is dependent on the wavelength of light, the distance between the slits, and the distance between the slits and the screen.

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Answer:

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2, A rectangulv Container of base socmby 30cm is filled
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Answers

Correct question is;

2, A rectangular Container of base 50 cm by 30cm is filled with water to be depth of 5cm. how much is the pressure exerted at the base? (take pw= 1000kg /m3 and g=10m/s)​

Answer:

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Answers

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[tex]ek = \frac{1}{2}(50)[/tex]

[tex]ek = 25j[/tex]

//

I'm not really sure but I do know that it's not 0 because the object is still moving, even if it's only moving at 1 m/s.

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Answers

Answer:

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Answers

Answer:

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Explanation:

Answer:

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Answers

Answer:

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