Question

In a Young's double slit experiment, slits are separated by $0.5mm$, and the screen is placed $150cm$ away. A beam of light consisting of two wavelengths, $650nm$ and $520nm$, 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. $15.6mm$
• B. $1.56mm$
• C. $7.8mm$
• D. $9.75mm$

Answer 1

The correct option is C $7.8mm$

$y=\frac{m\times 650\times {10}^{-9}\times D}{d}=\frac{n\times 520\times {10}^{-9}\times D}{d}$
$\Rightarrow \frac{m}{n}=\frac{4}{5}\Rightarrow$ minimum values of m and n will be 4 and 5 respectively.
$y=\frac{4\times 650\times {10}^{-9}\times 1.5}{5\times {10}^{-4}}meter=7.8mm$