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

Answer 1

Answer: (B)

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