Question

In a Young's double-slit experiment, $\lambda =500nm,d=1mm$, and $D=1m$. The minimum distance from the central maximum for which the intensity is half of the maximum intensity is

• A. $2\times {10}^{-4}m$
• B. $1.25\times {10}^{-4}m$
• C. $4\times {10}^{-4}m$
• D. $2.5\times {10}^{-4}m$

Answer 1

The correct option is B $1.25\times {10}^{-4}m$

Let $x$ be the minimum distance and $I_o$ be the intensity of central maximum then,

$\frac{I_o}{2}=I_o{\cos }^2\left(\frac{\pi x}{\beta }\right)$

$\frac{\pi x}{\beta }=\frac{\pi }{4}\Rightarrow x=\frac{\lambda D}{4d}=\frac{5\times {10}^{-7}\times 1}{4\times {10}^{-3}}=1.25\times {10}^{-4}m$