Question 294: Waves Optics

Question

In a diffraction pattern due to single slit of width ' $a$ ', the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \mathop A\limits^o$ is incident on the slit. The first secondary maximum is observed at an angle $\left[\sin 30=\frac{1}{2}\right]$

Accepted Answer

In the context of a single-slit diffraction pattern, the first minimum occurs when the light of wavelength $5000 \, 	ext{ÃÂ}$ is diffracted at an angle of $30^\circ$. This scenario arises from the relationship for minima in single-slit diffraction as described by:
$ a \sin 	heta_n = n \lambda $
For the first minimum ($n=1$), we can express it as:
$ a \sin 30^\circ = \lambda \quad 	ext{(Equation i)} $
Given the value of $\sin 30^\circ = \frac{1}{2}$, we substitute to get:
$ a \cdot \frac{1}{2} = \lambda $
Next, to find the angle for the first secondary maximum, we consider:
$ a \sin 	heta_n = (2n + 1) \frac{\lambda}{2} $
For the second secondary maximum ($n=1$), the equation becomes:
$ a \sin 	heta_n = (2 \cdot 1 + 1) \frac{\lambda}{2} = \frac{3\lambda}{2} \quad 	ext{(Equation ii)} $
By dividing Equation i by Equation ii, we get:
$ \frac{\frac{1}{2}}{\sin 	heta_n} = \frac{2}{3} $
Solving for $\sin 	heta_n$, we find:
$ \sin 	heta_n = \frac{3}{4} $
Thus, the angle $	heta_n$ for the first secondary maximum is:
$ 	heta_n = \sin^{-1}\left(\frac{3}{4}ight) $

Answer

$\sin ^{-1}\left(\frac{1}{2}\right)$

Answer

$\sin ^{-1}\left(\frac{3}{4}\right)$

Answer

$\sin ^{-1}\left(\frac{1}{4}\right)$

Answer

$\sin ^{-1}\left(\frac{3}{5}\right)$

Waves Optics

In a diffraction pattern due to single slit of width ' a ', the first minimum is observed at an angle 30^{\circ} when light of wavelength 5000 \mathop A\limits^o is incident on the slit. The first secondary maximum is observed at an angle \left[\sin 30=\frac{1}{2}\right]

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