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**Hint**To solve this question, the formula for the fringe width is to be used. The linear aperture can be treated as the single slit of the Young’s single slit diffraction experiment, whose width is given in the question. Also, the screen must be placed at the focus of the lens so as to obtain the diffraction pattern. So the distance of the screen from the slits is equal to the focal length of the lens. On substituting these values into the formula for the fringe width, we will get the final answer.

**Formula used:**The formula used is

$\beta = \dfrac{{\lambda D}}{d}$, here $\beta $ is the width of the bright or the dark fringe, $D$ is the distance of the screen from the slits of width $d$ and ]$\lambda $ is the wavelength of the light.

**Complete step-by-step solution:**

From the given information, we can say that we have a single slit diffraction experimental set up, where the linear aperture is nothing but the single slit. According to the question, the aperture is placed just in front of the lens and a parallel beam is incident on the aperture. We know that when a parallel beam of light strikes a lens, the image is formed at the focus. Since the distance between the aperture and the lens is negligible, the screen must be placed at the focus of the lens so that the distance of the screen from the slit is equal to the focal length of the lens. This means, we have

$D = 60cm$.................(1)

$d = 0.02cm$...............(2)

We know that the distance between a bright fringe and a dark fringe is equal to the fringe width. Therefore, the distance of the first dark band from the centre of the screen is equal to the fringe width which is given by

$\beta = \dfrac{{\lambda D}}{d}$

Substituting (1) and (2)

$\beta = \dfrac{{\lambda \left( {60} \right)}}{{0.02}}$

Also, according to the question, the wavelength is given as $\lambda = 5 \times {10^{ - 5}}cm$. Putting it above, we get

\[\beta = \dfrac{{\left( {5 \times {{10}^{ - 5}}} \right)\left( {60} \right)}}{{0.02}}cm\]

On solving we finally get

$\beta = 0.15cm$

**Hence, the correct answer is option A.**

**Note:**We do not need to convert the units of the wavelength, the focal length, and the width given in the question in meters. This is because all of these are given in centimeters and the values in the options are also in centimeters.

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