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monitta
3 years ago
14

In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen i

s D=48.0 cm away from the slits. What is the linear distance Δx between the eighth order maximum and the fourth order maximum on the screen?
Physics
1 answer:
pantera1 [17]3 years ago
5 0

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

<u>Δx = 4.68 x 10⁻³ m = 4.68 mm</u>

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