Question

Young's double slit experiment is performed at $589$ nm light with a distance of $2.00$ m between the slits and the screen. The tenth interference minimum is observed $7.26$ mm from the central maximum. The spacing of the slits is

• A. $15.4mm$
• B. $154mm$
• C. $1.54mm$
• D. $0.154mm$

Answer 1

The correct option is C $1.54mm$

${10}^{th}mimima=\frac{19}{2}\frac{\lambda D}{d}$
$7.26\times {10}^{-3}=\frac{19}{2}\times \frac{589\times {10}^{-9}\times 2}{d}$
$d=\frac{19\times 589\times {10}^{-9}\times 2}{2\times 7.26\times {10}^{-3}}$
$=1540\times {10}^{-6}$
$=1.54\times {10}^{-3}m$
$=1.54mm$