In Young's double slit interference experiment, the slit widths are...

We know that, $\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{\omega_1}{\omega_2}+1}\right)^2}{\left(\sqrt{\frac{\omega_1}{\omega_2}-1}\right)^2}I_{\max }$ and $I_{\min }$ are maximum and minimum intensity $\omega_1$ and $\omega_2$ are widths of two slits \therefore \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{1}{25}}+1\right)^2}{ \left(\sqrt{\frac{1}{25}}-1\right)^2}\left(\frac{\omega_1}{\omega_2}=\frac{1}{25} \text{ given }\right) On solving we get, \frac{I_{\max }}{I_{\min }}=\frac{\frac{36}{25}}{\frac{16}{25}}=\frac{9}{4}=$$9: 4