Question

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30° in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is ____ $\mu$m.

Hint:

In single-slit diffraction, the angular separation between adjacent minima is used to calculate the width of the slit. Ensure that you convert the wavelength to meters and carefully solve for the slit width.

Answer 1

Correct Answer: 6

\begin{center} \includegraphics[width=0.25\linewidth]{q48_digram.png} \end{center} The angular separation for the minima in a single-slit diffraction is given by: \[ \theta_1 = \sin^{-1}\left( \frac{2\lambda}{a} \right), \quad \theta_2 = \sin^{-1}\left( \frac{3\lambda}{a} \right) \] where \( \lambda = 628 \, \text{nm} \) is the wavelength and \( a \) is the slit width. Also, we know: \[ \theta_1 + \theta_2 = 30^\circ \] \[ \Rightarrow \sin^{-1}\left( \frac{2\lambda}{a} \right) + \sin^{-1}\left( \frac{3\lambda}{a} \right) = \frac{\pi}{6} \] Solving this, we find: \[ a = 6.07 \, \mu m \] Thus, the width of the slit is \( a = 6 \, \mu m \).