Question

In a young’s double-slit experiment $\lambda$ = 500 nm, d = 1m 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

As, $I=4I_{0}co{s}^2\frac{\varphi }{2}$
We will get, $\varphi =\frac{\pi }{2}$ for $I=\frac{I_0}{2}$
Hence, path difference $=\frac{\lambda }{4}$
So, distance y $=\frac{D\left(\lambda /4\right)}{d}$

Placing the values given, we will get y = $1.25\times {10}^{-4}$m