#objective focal length

Optical Instruments - Simple and Compound Microscope

A telescope has an objective focal length of \( 150 \, \text{cm} \) and an eyepiece focal length of \( 5 \, \text{cm} \)

**Microscope principle** objective of short focal length forms real image between F_e and 2F_e of eyepiece, eyepiece magnifies to virtual image at D, total magnification product, for f_o=2 cm, f_e=5 cm, L=20 cm, m_o≈10, M_e≈6, M≈60, illustrating high magnification from two stages. Tube length = f_o + f_e . f_o = 150 cm , f_e = 5 cm . Tube length = 150 + 5 = 155 cm . Substituting values gives 155 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

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

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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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Optical Instruments - Simple and Compound Microscope

A compound microscope has an objective of focal length \( 2 \, \text{cm} \) and eyepiece of focal length \( 5 \, \text{c

**Simple microscope** magnification M = 1 + D/f when image at D=25 cm near point, M = D/f when image at infinity (relaxed eye), f focal length (cm), D least distance of distinct vision 25 cm. For f=4 cm, image at 25 cm, M=1+25/4=7.25, angular magnification ratio of angle subtended by image to that by object at D. Objective magnification: m_o = (L/f_o) = (18/2) = 9 . Eyepiece magnification: m_e = (D/f_e) = (25/5) = 5 . Total magnification: m = m_o × m_e = 9 × 5 = 45 . Substituting values gives 45, which matches expected image position and magnification from mirror/lens

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

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Thin Lenses - Lens Formula, Magnification and Power

A telescope has an objective of focal length \( 160 \, \text{cm} \) and an eyepiece of focal length \( 8 \, \text{cm} \)

**Convex lens image formation**: object beyond 2F (u>2f) real inverted diminished between F and 2F, at 2F same size at 2F, between F and 2F magnified beyond 2F, at F image at infinity, within F virtual erect magnified same side. For f=15 cm, u=30 cm=2f, m = v/u =30/30=1? Actually v=30 cm, m=-1, same size inverted. Magnifying power: m = (f_o/f_e) . f_o = 160 cm , f_e = 8 cm . m = (160/8) = 20 . Substituting values gives 20, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror),

Ref: NCERT > Physics Book > Ray Optics > Thin Lenses - Lens Formula, Magnification and Power

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