Question

A Young's interference experiment is performed with monochromatic light. The separation between the slits is $0.5mm$, and the interference pattern on a screen is $3.5m$ away, shows the first order maximum at $3.6mm$ from the centre of the pattern. The wavelength is

• A. $515nm$
• B. $315mm$
• C. $215cm$
• D. $15m$

Answer 1

The correct option is A $515nm$

$d=0.5mm=0.5\times {10}^{-4}m$
$D=3.5m$
$y_1=3.6\times {10}^{-4}m$
$y_1=\frac{D\lambda }{d}$
$\Rightarrow \lambda =\frac{d\times y_1}{D}=\frac{0.5\times {10}^{-4}\times 3.6\times {10}^{-4}}{3.5}=0.514\times {10}^{-8}m=5.14nm$