JEE PYQ: Wave Optics Question 36
Question 36 - 2019 (10 Jan Shift 2)
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 ($S_1$)?
(1) $\frac{\lambda}{2(\sqrt{5} - 2)}$
(2) $\frac{\lambda}{(\sqrt{5} - 2)}$
(3) $\frac{\lambda}{2(5 - \sqrt{2})}$
(4) $\frac{\lambda}{(5 - \sqrt{2})}$
Show Answer
Answer: (1)
Solution
Here, $x_1 = 2d$ and $x_2 = \sqrt{5}d$
For first minima, $\Delta x = \frac{\lambda}{2}$
$\therefore \Delta x = x_2 - x_1 = \sqrt{5}d - 2d = \frac{\lambda}{2}$
$\Rightarrow d = \frac{\lambda}{2(\sqrt{5} - 2)}$
BETA
