JEE Main · 2017 · 02 Apr 2017 · Offline · Wave Optics

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In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

(A)9.75 mm
(B)15.6 mm
(C)1.56 mm
(D)7.8 mm✓ Correct

Answer: (D)

Solution

Each colour has its own fringe width

β = \frac{λ D}{d}

: for

650

nm,

β_{1} = 1.95

mm, and for

520

nm,

β_{2} = 1.56

mm. Bright fringes coincide where

n_{1} β_{1} = n_{2} β_{2}

, i.e.

\frac{n _{1}}{n _{2}} = \frac{1.56}{1.95} = \frac{4}{5}

, first satisfied at

n_{1} = 4, n_{2} = 5

. The coincidence then sits at

4 \times 1.95 = 7.8

mm from the central maximum.

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option

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