Question

In Young's double slit experiment, the slits are $2$ mm apart and are illuminated by photons of two wavelengths ${\lambda }_1=12000\stackrel{o}{A}$ and ${\lambda }_2=10000\stackrel{o}{A}$. At what minimum distance from the common central bright fringe on the screen $2$ m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

• A. $3$ mm
• B. $8$ mm
• C. $6$ mm
• D. $4$ mm

Answer 1

The correct option is C $6$ mm

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{h_1}{h_2}$

$=\frac{12000}{10000}=\frac{6}{5}$

$=\frac{u_1{\lambda }_{1}D}{d}=\frac{5\times 12000\times {10}^{-10}\times 2}{2\times {10}^{-3}}$

$=5\times 1.2\times {10}^4\times {10}^{-10}\times {10}^3=6mm$