Question

A beam of light consisting of two wavelengths 6500 A and 5200 A, is used to obtain interference fringes in a Young's double-slit experiment.
(i) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 6500 A.
(ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.

Solution

Given,
Wavelength,

λ_{1}

= 6500

\overset{o}{A}

Wavelength,

λ_{2}

= 5200

\overset{o}{A}

(i) The distance of the mth bright fringe from the central maximum is given by

y_{m} = \frac{mλD}{d}

∴

y_{3} = \frac{3 λD}{d} = \frac{3 \times (6500 \times 10^{- 10}) \times 1.20}{2 \times 10^{- 3}}

= 1.17 \times 10^{- 3} m = 1.17 mm

(ii) Let, the nth bright fringe of wavelength λ_n and the mth bright fringe of wavelength λ_mcoincide at a distance y from the central maximum, then

y = \frac{{mλ}_{m} D}{d} = \frac{{nλ}_{n} D}{d}

i.e.,

\frac{m}{n} = \frac{λ_{n}}{λ_{m}} = \frac{6500}{5200} = \frac{5}{4}

The least integral value of m and n which satisfy the above are m = 5 and n = 4

i.e., the 5th bright fringe of wavelength 5200

\overset{0}{A}

coincides with the 4th bright fringe of wavelength 6500

\overset{o}{A}

.

The smallest value of y at which the bright fringes coincides is,

y_{\min} = \frac{{mλ}_{m} D}{d} = \frac{5 \times (5200 \times 10^{- 10}) \times 1.20}{2 \times 10^{- 3}} = 1.56 \times 10^{- 3} m = 1.56 mm .

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