Practice question

**Lens combination** effective focal length calculation uses reciprocal sum, for f₁=10 cm, f₂=20 cm in contact, 1/F=1/10+1/20=3/20, F=6.67 cm, P=15 D, stronger converging than either alone, illustrating power addition. In a simple microscope, placing the object between the lens and focal point results in a virtual, erect, and magnified image. The closer the object is to the lens (inside F), the greater the divergence of rays, increasing the apparent size of the virtual image seen by the observer. Substituting values gives Due to increased divergence of rays, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f