b
We know in displacement method,

\(m _1=\frac{ v }{ u }\) and

\(m _2=\frac{ u }{ V }\)

Thus, \(m_1 \times m_2=1\)

\(m _1- m _2=\frac{ v }{ u }-\frac{ u }{ v }\)

\(m _1- m _2=\frac{ V ^2- u ^2}{ uv }\)

From the figure,

Given, \(v-u=x\)

\(\Rightarrow m _1- m _2=\frac{( v + u )( v - u )}{ uv }=\frac{( v + u ) x }{ uv }\)

Using lens formula,

\(\frac{1}{f}=\frac{1}{V}-\frac{1}{u}\)

For Position 1, we have

Object Distance \(=- u\)

Image Distance \(=v\)

\(\frac{1}{f}=\frac{1}{v}-\frac{1}{-u}\)

\(\frac{1}{f}=\frac{u+v}{u v}\)

\(\Rightarrow m _1- m _2=\frac{( v + u ) x }{ uv }=\frac{1}{ f } \times x\)

\(\Rightarrow m _1- m _2=\frac{ x }{ f }\)

\(f=\frac{x}{m_1-m_2}\)