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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 4.60 mm away from the slits.

A) Which laser has its first maximum closer to the central maximum?

B) What is the distance between the first maxima (on the same side of the central maximum) of the two patterns?

Sagot :

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Answer:

A. Laser 1 has its first maximum closer to the central maximum.

B. 0.077 mm

Explanation:

A) Which laser has its first maximum closer to the central maximum?

Since for a double slit interference, sinθ = mλ/d where θ is the angle between the central axis and the maximum point, m = order of fringe, λ = wavelength of light and d = width of slit.

Also tanθ = y/D where θ is the angle between the central axis and the maximum point, y = distance between maximum point and central axis and D = distance between slit and screen = 4.60 mm.

Since θ is small,  sinθ ≅  θ and  tanθ ≅ θ

So, sinθ ≅ tanθ  

mλ/d = y/D

So, y = mλD/d for the first maximum, m = 1.

So,  y = (1)λD/d

y = λD/d

For the Laser 1, λ = d/20

So, y = λD/d

y = (d/20)D/d  

y = D/20

y = 4.60 mm/20

y = 0.23 mm

For the Laser 2, λ' = d/15

So, y' = λ'D/d

y' = (d/15)D/d  

y' = D/15

y' = 4.60 mm/15

y' = 0.307 mm

Since y = 0.23 mm < y' = 0.307 mm, Laser 1 has its first maximum closer to the central maximum.

B) What is the distance between the first maxima (on the same side of the central maximum) of the two patterns?

The distance between the first maxima (on the same side of the central maximum) of the two patterns is Δy = y' - y = 0.307 mm - 0.23 mm = 0.077 mm

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