In a Young’s interference experiment, the two slits are separated by 0.150 mm and the incident light includes two wavelengths: l1 5 540 nm (green) and l2 5 450 nm (blue). The overlapping interference patterns are observed on a screen 1.40 m from the slits. Calculate the minimum distance from the center of the screen to a point where a bright fringe of the green light coincides with a bright fringe of the blue light:

2.52 × 10⁻² cm

Explanation:

The distance of bright fringe from the center of the screen is given by the formula

Here, wavelength is λ, Distance of the screen from the slits is D, seperation between the

slits is d.

Separation between the slits, d = 0.15 mm

Distance of the screen from the slits = 1.40 m

We have a wavelength, λ1 = 540 nm

                                           =

By substituting all these values in the above equation we get

                y1 = mλD/d

We have a wavelength, λ2 = 450 nm

                                           =

By substituting all these values in the above equation we get

According to the problem, these two distance are coincides with each other.

So,

by testing values, the above equation is satisfied only when, m = 5 and m' = 6

Then from the above we have

                           y1 = y2 = 0.0252 m

                                         = 2.52 × 10⁻² cm

b.

explanation:

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magnified images possible images can be real = projected onto a screen