Angular width of the central maxima in the Fraunhoffer diffraction...

The angular width of central maxima is given by 2 $\theta=\frac{2 \lambda}{\mathrm{a}}$where, $\lambda$ = wavelength of light used $\mathrm{a}=$ width of the slit For $\lambda_1=6000 Å, 2 \theta=\theta_0$ (given) For another light of wavelength $\lambda_2$ (says), the angular wid th decreases by $30 \%$ so, 2 \theta=\left(\frac{100-30}{100}\right) \theta_0=\frac{70}{100} \theta_0=0.7 \theta_0 As slit width is constant, so using Eq. (i) for these values, we get $\begin{gathered}\frac{\theta_0}{0.7 \theta_0}=\frac{\lambda_1}{\lambda_2}\\ \Rightarrow \lambda_2=\lambda_1 \times0.7=6000 \times0.7=4200 Å \end{gathered}