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#total magnification

2 public questions tagged with this topic.

Combination of Lenses and Lens Systems

A compound microscope has an objective of focal length \( 2 \, \text{cm} \) and tube length \( 20 \, \text{cm} \). If th

**Two lenses in contact** behave as single lens with power sum, magnification product m = m₁×m₂. For f₁=20 cm (P₁=5 D), f₂=-20 cm (P₂=-5 D) in contact, P=0, F infinite, afocal system, beam emerges parallel, principle of corrective lenses for myopia/hypermetropia. Objective magnification: m_o = (L/f_o) . L = 20 cm , f_o = 2 cm ⇒ m_o = (20/2) = 10 . Eyepiece magnification: m_e = 5 (given). Total magnification: m = m_o × m_e = 10 × 5 = 50 . Substituting values gives 50, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f

Ref: NCERT > Physics Book > Ray Optics > Combination of Lenses and Lens Systems

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Optical Instruments - Simple and Compound Microscope

A compound microscope has an objective of focal length \( 1.5 \, \text{cm} \) and tube length \( 18 \, \text{cm} \). If

**Compound microscope** M = m_o × M_e, objective magnification m_o = v_o/u_o ≈ L/f_o, L tube length, eyepiece M_e = 1+D/f_e (image at D) or D/f_e (infinity). Objective forms real inverted magnified image at focal plane of eyepiece, eyepiece acts as simple microscope magnifying it. Objective magnification: m_o = (L/f_o) = (18/1.5) = 12 . Eyepiece magnification: m_e = (D/f_e) = (25/6) ≈ 4.17 . Total magnification: m = m_o × m_e = 12 × 4.17 ≈ 50 . Substituting values gives 50, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/

Ref: NCERT > Physics Book > Ray Optics > Optical Instruments - Simple and Compound Microscope

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