Q129 / 180 · NEET 2016 (Phase 1) · Physics

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 $5000 Å$ is incident on the slit. The first secondary maximum is observed at an angle of:

A
$\sin^{- 1} (\frac{1}{4})$
B
$\sin^{- 1} (\frac{2}{3})$
C
$\sin^{- 1} (\frac{1}{2})$
Correct answer
D
$\sin^{- 1} (\frac{3}{4})$

Solution

$a \sin 30 = λ$

$a \sin θ = \frac{3 λ}{2}$

$\frac{\sin θ}{\sin 30} = \frac{3}{2}$

$\sin θ = \frac{3}{2} \times \frac{1}{2}$

$\sin θ = \frac{3}{4}$

$θ = \sin (\frac{3}{4})$

Hence, the correct answer is option: 'D'.

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In a diffraction pattern due to a single slit of width a the first minimum is observed at an angle 30o when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of:

Solution

More Wave Optics Questions

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 $5000 Å$ is incident on the slit. The first secondary maximum is observed at an angle of: