A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit.
CBSE OUTSIDE DELHI - SET 1 2013
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Hint: From condition of diffraction, $\sin\theta = \text{n}\lambda\text{ (for minima) }$ $ = \bigg(\text{n} + \frac{1}{2}\bigg)\lambda\text{ for maxima}$ Provided n =1, 2, 3... and n =0 for central maxima,From condition of minima
$\text{a}\sin\theta = \lambda$ (n =1)
Since the value of $\lambda$ is of nm, so $\text{a},\theta = \lambda$ $\text{a}.\frac{\text{y}}{\text{D}} = \lambda$ $\bigg[\text{angle} = \frac{\text{arc}}{\text{radius}}\bigg]$ $\therefore\text{a} = \frac{\lambda\text{D}}{\text{y}}$ $\text{a} =\frac{500\times10^{-9}\times1}{2.5\times10^{-3}}\text{m} =2x10^{-4}\ m.$
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