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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 4.80m away from the slits.Part AWhich laser has its first maximum closer to the central maximum?Part BExpress your answer in meters.Part CWhat is the distance \Delta _y__max-max_ between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?Express your answer in meters.

Respuesta :

Answer:

a) that laser 1 has the first interference closer to the central maximum

c) Δy = 0.64 m

Explanation:

The interference phenomenon is described by the expression

         d sin θ = m λ

Where d is the separation of the slits, λ the wavelength and m an integer that indicates the order of interference

For the separation of the lines we use trigonometry

        tan θ = sin θ / cos θ = y / x

In interference experiments the angle is very small

          tan θ = sin θ = y / x

         d y / x = m λ

a) and b) We apply the equation to the first laser

          λ = d / 20

          d y / x = m d / 20

          y = m x / 20

          y = 1 4.80 / 20

          y = 0.24 m

The second laser

        λ = d / 15

          d y / x = m d / 15

          y = m x / 15

          y = 0.32 m

We can see that laser 1 has the first interference closer to the central maximum

c) laser 1

They ask us for the second maximum m = 2

            y₂ = 2 4.8 / 20

            y₂ = 0.48 m

For laser 2 they ask us for the third minimum m = 3

In this case to have a minimum we must add half wavelength

         y₃ = (m + ½) x / 15

         m = 3

         y₃ = (3 + ½) 4.8 / 15

         y₃ = 1.12 m

        Δy = 1.12 - 0.48

        Δy = 0.64 m

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