MCQ
In Young’s double slit experiment, the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to
  • A
    $\frac{4}{3}\left( {\frac{{{\lambda _2}}}{{{\lambda _1}}}} \right)$
  • $\frac{4}{3}\left( {\frac{{{\lambda _1}}}{{{\lambda _2}}}} \right)$
  • C
    $\frac{3}{4}\left( {\frac{{{\lambda _2}}}{{{\lambda _1}}}} \right)$
  • D
    $\frac{3}{4}\left( {\frac{{{\lambda _1}}}{{{\lambda _2}}}} \right)$

Answer

Correct option: B.
$\frac{4}{3}\left( {\frac{{{\lambda _1}}}{{{\lambda _2}}}} \right)$
b
(b) Position of nth maxima from central maxima is given by ${x_n} = \frac{{n\lambda D}}{d}$
==> ${x_n} \propto n\lambda $ ==> $\frac{{{d_1}}}{{{d_2}}} = \frac{{{n_1}{\lambda _1}}}{{{n_2}{\lambda _2}}} = \frac{{8{\lambda _1}}}{{6{\lambda _2}}} = \frac{4}{3}\left( {\frac{{{\lambda _1}}}{{{\lambda _2}}}} \right)$

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