Question

When light of wavelength 5000Å is used in the single slit diffraction experiment,the first diffraction minima is formed at the position θ. If the width of the slit is 10^-4 cm,the magnitude of θ is: a)30 b)45 c)60 d)15.

Answer 1

Answer on Question #38004 – Physics – Other

Question: when light of wavelength $5000\,\mathrm{\AA}$ is used in the single slit diffraction experiment, the first diffraction minimum is formed at the position $\theta$. If the width of the slit is $10^{-4}\,\mathrm{cm}$, the magnitude of $\theta$ is: $a)30$; $b)45$; $c)60$; $d)15$.

Solution: in the single slit diffraction experiment the condition for the minimum points is

$$a \cdot \sin \theta = n \cdot \lambda, n = \pm 1, \pm 2, \dots$$

For the first diffraction minimum we obtain $a \cdot \sin \theta = \lambda \rightarrow \sin \theta = \frac{\lambda}{a}$.

$$\sin \theta = \frac{\lambda}{a} = \frac{5000 \cdot 10^{-10}}{10^{-6}} = 0,5.$$

The solution of the equation $\sin \theta = 0,5$ is $\theta = 30{}^\circ$.

Answer: $a)30{}^\circ$.