Concept:For a combination of lenses in contact, the equivalent focal length is given by feq1=f11+f21. For lenses with a gap, treat each lens separately; the image from the first lens becomes the object for the second lens, applying the lens formula to each.Explanation:Given convex lens f1=+5 cm and concave lens f2=−4 cm. Object is placed u=−10 cm from the convex lens.Case 1: No gap (d=0).Equivalent focal length: feq1=51+−41=−201, so feq=−20 cm.Using lens formula: v1−u1=feq1.v1−−101=−201⇒v1=−201−101=−203, so v=−320 cm.Magnification m1=uv=−10−20/3=32.Case 2: Gap d=1 cm.For the first lens: v11−−101=51⇒v1=+10 cm.Magnification mL1=u1v1=−1010=−1.The image from first lens is 10 cm to its right; the second lens is 1 cm to the right, so object distance for second lens u2=v1−d=10−1=+9 cm (virtual object).For the second lens: v21−91=−41⇒v21=−41+91=−365, so v2=−536 cm.Magnification mL2=u2v2=9−36/5=−54.Total magnification m2=mL1×mL2=(−1)×(−54)=54.Required ratio: $\left| \frac{m_1}{m_2} \right| = \frac{2/3}{4/5} = \frac{2}{3} \times \frac{5}{4} = \frac{10}{12
