Adouble slit apparatus is held 1.2 m from a screen. /4] (a) when red light (λ = 600 nm) is sent through the double slit, the interference pattern on the screen shows a distance of 12.5 cm between the first and tenth dark fringes. what is the separation of the slits? (b) what will be the difference in path length for the waves travelling from each slit to the tenth nodal line?

A. To solve for part A, we use the formula of Young’s double slit equation:

x = λ L m / d

Where,

x = distance between adjacent dark lines or fringes = 12.5 cm = 0.125 m

λ = wavelength of light = 600 × 10^-9 m

L = distance from the two sources of light to the screen = 1.2 m

m = number of fringes = 10 (tenth) – 1 (first) = 9

d = separation of slits = unknown

Rearranging the equation in terms of d and plugging in the values:

d = λ L m / x

d = (600 × 10^-9 m) (1.2 m) (9) / 0.125 m

d = 5.184 × 10^-5 m

d = 51.84 μm

B. The formula for path difference is:

path difference = (2 m + 1) (λ / 2)
path difference = (2 * 9 + 1) (600 × 10^-9 m / 2)

b

explanation:

bc its just b

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