Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer 1

Answer:

Answer:

a)Δy = 81.7mm

b)Δy = 32.7cm

Explanation:

To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:

$Tan \theta = (y)/(D)$

where D is the separation between the slits and the screen where the interference pattern is observed.

a) In this case:

Δy = |y1max (λ1) − y1max (λ2)|

Δy = $|(D\lambda _1)/(d) - (D\lambda _2)/(d) |$

Δy = $D |(d/20)/(d) - (d/15)/(d) |$

Δy = $D |(1)/(20) - (1)/(15) |$

Δy = $4.90 |(1)/(20)- (1)/(15) |$

Δy = 81.7mm

The separation between these maxima is 81.7 mm

Δy = |y₂max (λ1) − y₂max (λ2)|

Δy = $D|(2(d/20))/(d) - (5(d/15))/(2d) |$

Δy = $4.90|(1)/(10) - (1)/(6) |$

Δy = 32.7cm

The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.

Answer 2

Answer:

Final answer:

We can solve the problem using the concepts of waveinterference and the formulas for maxima and minima positions (i.e., y = L*m*λ/d and y = L*(m+1/2)*λ/d respectively). The difference between the first maxima of the two patterns is 4.9/60 m and the difference between the second maximum of laser 1 and the third minimum of laser 2 is also 4.9/60 m.

Explanation:

The problem described deals with wave interference and can be addressed using the formulas for path difference and phasedifference.

To answer part a, we need to find the difference between the positions of the first maxima for the two lasers. The position of any maxima in an interference pattern can be found using the formula: y = L * m * λ / d, where L is the distance from the slits to the screen, m is the order of the maxima, λ is the wavelength, and d is the slit separation.

So for the first laser (λ1=d/20) the position of the first maxima would be y1 = 4.9m * 1 * (d/20) / d =4.9/20 m.

And for the second laser (λ2 = d/15) the position of the first maxima would be y2= 4.9m * 1 * (d/15) / d =4.9/15 m.

Then, the distance Δ ymax-max between the first maxima of the two patterns is y2-y1= 4.9/15 m - 4.9/20 m = 4.9/60 m.

Answering part b involves finding the positions of the second maximum of laser 1 and the third minimum of laser 2. The position of any minimum in an interference pattern can be calculated using the formula: y = L * (m+1/2) * λ / d. For the second maximum of laser 1, we have y1max2 = 4.9 m * 2 * (d/20) / d = 4.9/10 m. For the third minimum of laser 2, we have y2min3 = 4.9m * (3.5) * (d/15)/d = 4.9*7/30 m. The difference Δymax-min is y2min3-y1max2= 4.9*7/30 m - 4.9/10 m = 4.9/60 m.

Learn more about Wave Interference here:

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Answers

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)

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Substituting Equation 2 into equation 1.

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making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

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Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

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Answers

Answer:

The answer is: Pressure increases linearly with the depth

Explanation:

In this case, the definition of pressure is:

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The minimum charge on any object cannot be less than

Answers

Answer:

1.6 x 10^{-19} Coulombs

Explanation:

In Physics, the standard unit of measurement of a charge is Coulombs and it's denoted by C. Also, the symbol for denoting a charge is Q.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

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The acrylic plastic rod is 20 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Eₚ = 2.70 GPa, vₚ = 0.4.

Answers

Given Information:

diameter = d = 15 mm

Length = L = 20 mm

Axial load = P = 300 N

Eₚ = 2.70x10⁹ Pa

vₚ = 0.4

Required Information:

Change in length = ?

Change in diameter = ?

Answer:

Change in length = 0.01257 mm

Change in diameter = -0.003772 mm

Explanation:

Stress is given by

σ = P/A

Where P is axial load and A is the area of the cross-section

A = 0.25πd²

A = 0.25π(0.015)²

A = 0.000176 m²

σ = 300/0.000176

σ = 1697792.8 Pa

The longitudinal stress is given by

εlong = σ/Eₚ

εlong = 1697792.8/2.70x10⁹

εlong = 0.0006288 mm/mm

The change in length can be found by using

δ = εlong*L

δ = 0.0006288*20

δ = 0.01257 mm

The lateral stress is given by

εlat = -vₚ*εlong

εlat = -0.4*0.0006288

εlat = -0.0002515 mm/mm

The change in diameter can be found by using

Δd = εlat*d

Δd = -0.0002515*15

Δd = -0.003772 mm

Therefore, the change in length is 0.01257 mm and the change in diameter is -0.003772 mm

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Answers

Answer:

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