Question

Consider a Young's double slit experiment as shown in figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $S_1$ ?

• A. $\frac{\lambda }{(5-\sqrt{2})}$
• B. $\frac{\lambda }{(\sqrt{2}-2)}$
• C. $\frac{\lambda }{2(5-\sqrt{2})}$
• D. $\frac{\lambda }{2(\sqrt{5}-2)}$

Answer 1

The correct option is D $\frac{\lambda }{2(\sqrt{5}-2)}$


Path difference on reaching $\text{P}$ from $S_1$ and $S_2$

$\Delta x=S_{2}P-S_{1}P=\sqrt{5}d-2d$

$\Delta x=(\sqrt{5}-2)d$

Also, for first minimum $\Delta x=\frac{\lambda }{2}$

$\Rightarrow (\sqrt{5}-2)d=\frac{\lambda }{2}$

$\Rightarrow d=\frac{\lambda }{2(\sqrt{5}-2)}$

Hence, option $\left(\text{D}\right)$ is correct.

Why this question ? Caution: In this question, we can't use formula $\Delta x=\frac{yd}{D}$ directly, because $d$ and $D$ are comparable, and this formula is valid for $(d<<D)$.