b
Here,  \(a=0.02\,cm\,=\,2 \times 10^{-4}\,m\)

\(\lambda=5 \times 10^{-5} \,\mathrm{cm}=5 \times 10^{-7}\, \mathrm{m}\)

\(D=60\, \mathrm{cm}=0.6\, \mathrm{m}\)

Position of first minima on the diffraction pattern

\({y_1} = \frac{{D\lambda }}{a} = \) \(\frac{{0.6 \times 5 \times {{10}^{ - 7}}}}{{2 \times {{10}^{ - 4}}}}\) \( = 15 \times {10^{ - 4}}{\text{m}} = 0.15\,\,{\text{cm}}\)