$(a)$ Determine the 'effective focal length' of the combination of two lenses,a convex lens of focal length $30 \; cm$ and a concave lens of focal length $20 \; cm$,if they are placed $8.0 \; cm$ apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
$(b)$ An object $1.5 \; cm$ in size is placed on the side of the convex lens in the arrangement $(a)$ above. The distance between the object and the convex lens is $40 \; cm$. Determine the magnification produced by the two-lens system,and the size of the image.

(N/A) Focal length of the convex lens,$f_{1} = 30 \; cm$.
Focal length of the concave lens,$f_{2} = -20 \; cm$.
Distance between the two lenses,$d = 8.0 \; cm$.
$(a)$ When the parallel beam of light is incident on the convex lens first:
Using the lens formula $\frac{1}{v_{1}} - \frac{1}{u_{1}} = \frac{1}{f_{1}}$ with $u_{1} = -\infty$,we get $v_{1} = f_{1} = 30 \; cm$.
This image acts as a virtual object for the concave lens. The distance $u_{2} = v_{1} - d = 30 - 8 = 22 \; cm$.
Applying the lens formula for the concave lens: $\frac{1}{v_{2}} - \frac{1}{22} = \frac{1}{-20} \implies \frac{1}{v_{2}} = \frac{1}{22} - \frac{1}{20} = \frac{10 - 11}{220} = -\frac{1}{220}$.
So,$v_{2} = -220 \; cm$. The parallel beam appears to diverge from a point $220 - 4 = 216 \; cm$ from the center of the combination.
When the parallel beam is incident on the concave lens first:
Using $\frac{1}{v_{2}} - \frac{1}{u_{2}} = \frac{1}{f_{2}}$ with $u_{2} = -\infty$,we get $v_{2} = f_{2} = -20 \; cm$.
This image acts as a real object for the convex lens. The distance $u_{1} = -(20 + 8) = -28 \; cm$.
Applying the lens formula for the convex lens: $\frac{1}{v_{1}} - \frac{1}{-28} = \frac{1}{30} \implies \frac{1}{v_{1}} = \frac{1}{30} - \frac{1}{28} = \frac{14 - 15}{420} = -\frac{1}{420}$.
So,$v_{1} = -420 \; cm$. The parallel beam appears to diverge from a point $420 - 4 = 416 \; cm$ from the center.
Since the results differ,the effective focal length is not a useful concept for this system.
$(b)$ For the convex lens,$u_{1} = -40 \; cm, f_{1} = 30 \; cm$. $\frac{1}{v_{1}} = \frac{1}{30} + \frac{1}{-40} = \frac{4-3}{120} = \frac{1}{120} \implies v_{1} = 120 \; cm$.
Magnification $m_{1} = \frac{v_{1}}{u_{1}} = \frac{120}{-40} = -3$.
For the concave lens,$u_{2} = 120 - 8 = 112 \; cm, f_{2} = -20 \; cm$. $\frac{1}{v_{2}} = \frac{1}{-20} + \frac{1}{112} = \frac{-112 + 20}{2240} = -\frac{92}{2240} \implies v_{2} \approx -24.35 \; cm$.
Magnification $m_{2} = \frac{v_{2}}{u_{2}} = \frac{-2240/92}{112} = -\frac{20}{92} \approx -0.217$.
Total magnification $m = m_{1} \times m_{2} = (-3) \times (-0.217) = 0.652$.
Size of image $h_{2} = m \times h_{1} = 0.652 \times 1.5 = 0.978 \; cm \approx 0.98 \; cm$.