MCQ
In a diffraction pattern due to a single slit of width $a$, the first minimum is observed at an angle $30^o $ when light of wavelength $5000 \;\mathring A$ is incident on the slit. The first secondary maximum is observed at an angle of
  • A
    $sin^{-1}$ $\left( {\frac{2}{3}} \right)$
  • B
    $sin^{-1}$$\left( {\frac{1}{2}} \right)$
  • $sin^{-1}$$\left( {\frac{3}{4}} \right)$
  • D
    $sin^{-1}$$\left( {\frac{1}{4}} \right)$

Answer

Correct option: C.
$sin^{-1}$$\left( {\frac{3}{4}} \right)$
c
For first minimum, the path difference between extreme waves,

$a \sin \theta=\lambda$

Here $\theta=30^{\circ} \Rightarrow \sin \theta=\frac{1}{2}$

$\therefore \quad a=2 \lambda$    ..... $(i)$

For first secondary maximum, the path difference between extreme waves

$a \sin \theta=\frac{3}{2} \lambda$ or $(2 \lambda) \sin \theta^{\prime}=\frac{3}{2} \lambda$  [Using eqn $(i)$]

or $\sin \theta^{\prime}=\frac{3}{4} \quad \therefore \theta^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)$

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