A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Distance of n^{th} fringe from central maxima, X_{n} = nλD/d

Where, λ is wavelength of light

D is distance from slits to screen

d is slit width

(a) We need to find distance of 3^{rd} from central maxima for wavelength 650 nm

n = 3

x_{3} = 3 × (650 nm) D/d

(b) The distance of n^{th} bright fringe from central maxima for two wavelengths, say λ_{1} and λ_{2}, are

X_{n} = nλ_{1}D/d

Y_{n} = nλ_{2}D/d

When bright fringe due to two wavelength coincide, their distance from central maxima is same

i.e., X_{n} = Y_{n}

n_{1}λ_{1}D/d = n_{2}λ_{2}D/d

n_{1}λ_{1} = n_{2}λ_{2}

n_{1}/n_{2} = λ_{2} /λ_{1}

= 520 / 650 = 8/10

Therefore, the bright fringe due to two wavelength coincide when n_{1} is integer multiple of 8 and n_{2} is integer multiple of 10.

Thus least distance for which they coincide,

S = n_{1} λ_{1}D/d

= 8 × (650 nm) D/d

[The question is not complete. This question cannot be answered completely without the values of d and D.]

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