First slide

Young's double slit Experiment

Question

A beam of light consisting of two wavelengths 650nm and 520nm is used to obtain interference fringes in Young's double slit experiment. The distance between slits is 2 mm and between the plane of slits and screen is 120 cm. The least distance from the central maximum where the bright hinges due to both wavelength coincide is

Easy

A

1.17 mm
B

3.34 mm
C

3.12 mm
D

1.560mm

Solution

let the n^th bright fringe of wavelength $λ_{m}$ and m ^th Bright fringe of wavelength $λ_{m}$ coincide at a distance x _nfrom central maximum .then
$x_{n} = \frac{{nλ}_{n} D}{2 d} = \frac{{mλ}_{m} D}{2 d} or \frac{m}{n} = \frac{λ_{n}}{λ_{m}} = \frac{650 nm}{520 nm} = \frac{5}{4} So, least integral values of m and n satisfying above requhemeDt are m = 5 andn = 4 Smallest value of x_{n}, is given by {(x_{n})}_{\min} = \frac{{nλ}_{n} D}{d} = \frac{4 \times (650 \times 10^{- 9}) \times 1.20}{2 \times 10^{- 3}} = 1.56 mm$

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