Optics 6 Question 38
40. White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between the slits is $b$ and the screen is at a distance $d(\gg>b)$ from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelengths are
(1984, 2M)
(a) $\lambda=b^{2} / d$
(b) $\lambda=2 b^{2} / d$
(c) $\lambda=b^{2} / 3 d$
(d) $\lambda=2 b^{2} / 3 d$
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Solution:
- At $P$ (directly infront of $\left.S _1\right) y=b / 2$
$\therefore$ Path difference,
$$ \Delta X=S _2 P-S _1 P=\frac{y \cdot(b)}{d}=\frac{\frac{b}{2}(b)}{d}=\frac{b^{2}}{2 d} $$
Those wavelengths will be missing for which
$$ \begin{aligned} \Delta X & =\frac{\lambda _1}{2}, \frac{3 \lambda _2}{2}, \frac{5 \lambda _3}{2} \ldots \\ \lambda _1 & =2 \Delta x=\frac{b^{2}}{d} \\ \lambda _2 & =\frac{2 \Delta x}{3}=\frac{b^{2}}{3 d} \\ \lambda _3 & =\frac{2 \Delta x}{5}=\frac{b^{2}}{5 d} \end{aligned} $$
