The relation between efficiency $(\eta)$ of Carnot engine and coefficient of performance ( $\eta_1$ ) of…

Relating efficiency and coefficient of performance

The efficiency of a Carnot engine operating between reservoirs at temperatures $T_1$ and $T_2$ is given by $\eta = 1 - \frac{T_2}{T_1}$, from which we obtain $\frac{T_2}{T_1} = 1 - \eta$.

The coefficient of performance for a refrigerator operating between the same reservoirs is $\eta_1 = \frac{T_2}{T_1 - T_2}$. Dividing numerator and denominator by $T_1$ gives $\eta_1 = \frac{T_2/T_1}{1 - T_2/T_1}$.

Substituting $T_2/T_1 = 1 - \eta$ yields $\eta_1 = \frac{1 - \eta}{1 - (1 - \eta)} = \frac{1 - \eta}{\eta}$.

Solving for $\eta$ gives $\eta_1\eta = 1 - \eta$, then $\eta(\eta_1 + 1) = 1$, resulting in $\eta = \frac{1}{\eta_1 + 1}$.

Final answer: $\boxed{A}$