MCQ
In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength ${\lambda _1}$ is found to be coincident with the third maximum at ${\lambda _2}$. So
  • A
    $3{\lambda _1} = 0.3{\lambda _2}$
  • B
    $3{\lambda _1} = {\lambda _2}$
  • ${\lambda _1} = 3.5{\lambda _2}$
  • D
    $0.3{\lambda _1} = 3{\lambda _2}$

Answer

Correct option: C.
${\lambda _1} = 3.5{\lambda _2}$
c
(c)Position of first minima = position of third maxima i.e., $\frac{{1 \times {\lambda _1}D}}{d} = \frac{{\left( {2 \times 3 + 1} \right)}}{2}\frac{{{\lambda _2}D}}{d}\,\, \Rightarrow \,{\lambda _1} = 3.5{\lambda _2}$

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