Given, width of linear aperture, $a = 0.02\text{ cm}$ Focal length of lens, $f = 60\text{ cm}$ Wavelength of light, $\lambda = 5 \times 10^{-5}\text{ cm}$ The distance of the $n^{\text{th}}$ dark band from the centre of the central maximum is given by: $y = \frac{n\lambda D}{a}$ Since the lens is placed immediately in front of the aperture, the distance of the screen $D$ is equal to the focal length $f$ of the lens. Thus, $D = f = 60\text{ cm}$. For the first dark band, $n = 1$: $y = \frac{1 \times 5 \times 10^{-5} \times 60}{0.02}$ $y = \frac{300 \times 10^{-5}}{2 \times 10^{-2}} = \frac{3 \times 10^{-3}}{2 \times 10^{-2}} = 1.5 \times 10^{-1}\text{ cm} = 0.15\text{ cm}$