\(\frac{-1}{120}-\frac{1}{40}=\frac{1}{ f }, \quad f =-30\,cm\)

Now,

\(\frac{-1}{ v ^2} dv -\frac{1}{ u ^2} du =-\frac{1}{ f ^2} df\)

Also \(dv = du =\frac{1}{20} cm\)

\(\frac{\frac{1}{20}}{(120)^2}+\frac{\frac{1}{20}}{(40)^2}=\frac{d f}{(30)^2}\)

On solving

\(df =\frac{1}{32}\,cm\)

\(k =32\)