a
Given \(\beta_1=0.2\,mm\)

Let wavelength corresponding to \(\beta\), is \(\lambda\) and \(d\) be distance b/w slits. We know, \(\beta=\frac{\lambda D}{d}\)

for \(\beta_1=0.2 mm =\frac{\lambda D}{\lambda}-\) (1)

Now, \(\lambda\) and \(d\) are invereased by \(10 \%\) i.e

\(\lambda^{\prime}=\frac{11}{10} \lambda \text { and } d^{\prime}=\frac{11}{10} d.\)

\(\therefore \quad \beta_2=\frac{\lambda^{\prime} D}{d^{\prime}}=\frac{11}{10} \lambda \times D \times \frac{10}{11 d}=\frac{\lambda D}{d}-2\)

from \((1)\) and \((2)\)

\(\beta_2=0.2\,mm\)