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 :

Question

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) 1.56 mm (B) 7.8 mm (C) 9.75 mm (D) 15.6 mm

Tardigrade Admin · Accepted Answer

For λ1 y =(mλ1D/d) For λ2 y =(nλ2D/d) ⇒ (m/n) = (λ2/λ1) = (4/5) For λ1 y = (mλ1D/d) , λ1 = 650 nm = 7.8 mm

1.56 mm

7.8 mm

9.75 mm

15.6 mm

For λ1​y=dmλ1​D​ For λ2​y=dnλ2​D​ ⇒nm​=λ1​λ2​​=54​ For λ1​ y=dmλ1​D​,λ1​=650nm =7.8mm

For $λ_{1} y = \frac{m λ _{1} D}{d}$
For $λ_{2} y = \frac{n λ _{2} D}{d}$
$\Rightarrow \frac{m}{n} = \frac{λ _{2}}{λ _{1}} = \frac{4}{5}$
For $λ_{1}$
$y = \frac{m λ _{1} D}{d}, λ_{1} = 650 nm$
$= 7.8 mm$