Question

In Young's double slit experiment, the screen is moved 30 cm towards the slits. As a consequence, the fringe width of the pattern changes by 0.09 mm. If the slits separation used is 2 mm, calculate the wavelength of light used in the experiment.

Hint:

\(\Delta\beta = \lambda\,\Delta D/d\), so \(\lambda = d\,\Delta\beta/\Delta D = 600\) nm.

Answer 1

Step 1: Note the data:
Screen shift \(\Delta D = 30\ \text{cm} = 0.30\ \text{m}\), change in fringe width \(\Delta\beta = 0.09\ \text{mm} = 9\times10^{-5}\ \text{m}\), and \(d = 2\ \text{mm} = 2\times10^{-3}\ \text{m}\).

Step 2: Use the fringe width formula:
The fringe width is \(\beta = \dfrac{\lambda D}{d}\). If \(D\) changes by \(\Delta D\) with \(\lambda\) and \(d\) fixed, the fringe width changes by
\[ \Delta\beta = \frac{\lambda\,\Delta D}{d} \]

Step 3: Find the wavelength:
\[ \lambda = \frac{d\,\Delta\beta}{\Delta D} = \frac{2\times10^{-3}\times 9\times10^{-5}}{0.30} = 6\times10^{-7}\ \text{m} \]
So \(\lambda = 600\ \text{nm}\).

Final Answer:
The wavelength of the light used is \[ \boxed{\lambda = 600\ \text{nm}} \]