Optics 6 Question 34
36. A light source, which emits two wavelengths $\lambda _1=400 nm$ and $\lambda _2=600 nm$, is used in a Young’s double-slit experiment. If recorded fringe widths for $\lambda _1$ and $\lambda _2$ are $\beta _1$ and $\beta _2$ and the number of fringes for them within a distance $y$ on one side of the central maximum are $m _1$ and $m _2$, respectively, then
(a) $\beta _2>\beta _1$
(2014 Adv.)
(b) $m _1>m _2$
(c) from the central maximum, 3rd maximum of $\lambda _2$ overlaps with 5 th minimum of $\lambda _1$
(d) the angular separation of fringes of $\lambda _1$ is greater than $\lambda _2$
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Solution:
- Fringe width $\beta=\frac{\lambda D}{d}$
$$ \begin{array}{lll} \text { or } & \beta \propto \lambda \\ \therefore & \lambda _2>\lambda _1 \\ \text { So } & \beta _2>\beta _1 \end{array} $$
Number of fringes in a given width
$$ \begin{aligned} & m=\frac{y}{\beta} \text { or } m \propto \frac{1}{\beta} \\ \Rightarrow \quad & m _2<m _1 \text { as } \beta _2>\beta _1 \end{aligned} $$
Distance of 3 rd maximum of $\lambda _2$ from central maximum
$$ =\frac{3 \lambda _2 D}{d}=\frac{1800 D}{d} $$
Distance of 5 th minimum of $\lambda _1$ from central maximum
$$ =\frac{9 \lambda _1 D}{2 d}=\frac{1800 D}{d} $$
So, 3 rd maximum of $\lambda _2$ will overlap with 5 th minimum of $\lambda _1$.
Angular separation (or angular fringe width) $=\frac{\lambda}{d} \propto \lambda$
$\Rightarrow$ Angular separation for $\lambda _1$ will be lesser.
