1. Let the initial lengths of brass and iron be: $L_b = 40\text{ cm}, \qquad L_i = ?$

2. The scale is made so that its effective length remains constant at $20\text{ cm}$ for all temperatures.

   Since the two strips expand differently, and the difference between their lengths is to remain constant, we require: $L_i - L_b = 20\text{ cm}$ initially, and this difference should not change with temperature.

3. After a temperature rise $\Delta T$:

   • Brass length becomes $L_b' = L_b(1+\alpha_b \Delta T)$
   • Iron length becomes $L_i' = L_i(1+\alpha_i \Delta T)$

   For the difference to remain constant, $L_i' - L_b' = L_i - L_b$

4. Substitute the expanded lengths: $L_i(1+\alpha_i \Delta T) - L_b(1+\alpha_b \Delta T) = L_i - L_b$

   Expanding, $L_i + L_i\alpha_i \Delta T - L_b - L_b\alpha_b \Delta T = L_i - L_b$

   Cancelling $L_i - L_b$ from both sides: $L_i\alpha_i \Delta T - L_b\alpha_b \Delta T = 0$

   $L_i\alpha_i = L_b\alpha_b$

5. Put the values: $L_i \times 1.2\times 10^{-5} = 40 \times 1.8\times 10^{-5}$

   $L_i = 40\times \frac{1.8}{1.2} = 40\times 1.5 = 60\text{ cm}$

6. Check the difference: $60 - 40 = 20\text{ cm}$ which matches the required scale length.

Therefore, the length of iron should be: $\boxed{60\text{ cm}}$