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STD 12 - 10. Wave optics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 10. Wave optics4 Marks
MCQ
Two coherent sources $S_1$ and $S_2$ are separated by a distance four times the wavelength $\lambda$ of the source. The sources lie along y axis whereas a detector moves along $+ x$ axis. Leaving the origin and far off points the number of points where maxima are observed is
  • A
    $2$
  • ✓
    $3$
  • C
    $4$
  • D
    $5$
✓

Answer

Correct option: B.
$3$
b
(b) From $\Delta {S_1}{S_2}D,$
${({S_1}D)^2} = {({S_1}{S_2})^2} + {({S_2}D)^2}$
${({S_1}P + PD)^2} = {({S_1}{S_2})^2} + {({S_2}D)^2}$
Here ${S_1}P$ is the path difference $ = n\lambda $ for maximum intensity.
$\therefore {(n\lambda + {x_n})^2} = {(4\lambda )^2} + {({x_n})^2}$
or ${x_n} = \frac{{16{\lambda ^2} - {n^2}{\lambda ^2}}}{{2n\lambda }}$
Then ${x_1} = \frac{{16{\lambda ^2} - {\lambda ^2}}}{{2\lambda }} = 7.5\,\lambda $
${x_2} = \frac{{16{\lambda ^2} - 4{\lambda ^2}}}{{4\lambda }} = 3\lambda $
${x_3} = \frac{{16{\lambda ^2} - 9{\lambda ^2}}}{{6\lambda }} = \frac{7}{6}\lambda $
${x_4} = 0$.
$\therefore $Number of points for maxima becomes $ 3$.

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