Discussion Forum

In 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

Answer:

Explanation:

let n₁th fringe formed due to first wavelength and n₂ th fringe formed due to second wavelength coincide i.e., their distance from common central maxima will be same

    i.e., Y n₁= Y n₂

  $\Rightarrow \frac{n_{1}\lambda_{1}D}{d}=\frac{n_{2}\lambda_{2}D}{d}$

$\Rightarrow \frac{n_{1}}{n_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{520}{650}=\frac{4}{5}$

Hence, distance of the point of coincidence from the central maxima is

   $y=\frac{n_{1}\lambda_{1}D}{d}=\frac{n_{2}\lambda_{2}D}{d}$

   = $\frac{4\times 650\times 10^{-9}\times15}{0.5\times 10^{-3}}=7.8nm$