Question

A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.30 m behind the slit. The distance between the first and second minima in the diffraction pattern is 3.95 mm .What is the width (in mm) of the slit?

Answer 1

The formula for the position of the \(m^{\text {th }}\) minimum from a single slit diffraction pattern is \(y=\frac{m \lambda D}{d}\)

The difference in \(y\) values between \(1^{\text {st }}\) and \(2^{\text {nd }}\) minima is \(\begin{aligned} y_{2}-y_{1} &=\left[\frac{2\left(633 \times 10^{-9} \mathrm{~m}\right)(1.30 \mathrm{~m})}{d}\right]-\left[\left(\frac{1\left(633 \times 10^{-9} \mathrm{~m}\right)(1.30 \mathrm{~m})}{d}\right)\right] \\ &=\frac{822.9 \times 10^{-9} \mathrm{~m}}{d} \end{aligned}\)

The difference of \(1^{\text {st }}\) and \(2^{\text {nd }}\) minima is equal to \(3.95 \mathrm{~mm}\). \(y_{2}-y_{1}=3.95 \mathrm{~mm}\)

\(\frac{822.9 \times 10^{-9} \mathrm{~m}}{d}=3.95 \mathrm{~mm}\)

The width of the slit is \(\begin{aligned} d &=\frac{\left(822.9 \times 10^{-9} \mathrm{~m}\right)}{\left(3.95 \times 10^{-3} \mathrm{~m}\right)} \\ &=0.208 \mathrm{~mm} \end{aligned}\)