MCQ
A parallel beam of monochromatic light is incident on a narrow rectangular slit of width $1\,mm$. When the diffraction pattern is seen on a screen placed at a distance of $2\,m$. the width of principal maxima is found to be $2.5\,mm$. The wave length of light is-$.............\mathring A$
  • $6250$
  • B
    $6200$
  • C
    $5890$
  • D
    $6000$

Answer

Correct option: A.
$6250$
a
(a)

Here the width of principal maxima is $2.5\,mm$, therefore its half width is $\frac{\beta}{2}=\frac{2.5}{2}=1.25 \times 10^{-3}\,m$

Diffraction angle $\theta=\frac{\beta / 2}{ D }=\frac{1.25 \times 10^{-3}}{2}$

$\therefore a \theta=\lambda \therefore \theta=\lambda / a =\frac{1.25 \times 10^{-3}}{2}$

$\lambda=\frac{1.25 \times 10^{-3}}{2} \times a =\frac{1.25 \times 10^{-3} \times 10^{-3}}{2}$

$\lambda=6.25 \times 10^{-7} m =6250 \mathring A$

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