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#eyepiece

19 public questions tagged with this topic.

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

**Astronomical telescope** in normal adjustment M = -f_o/f_e, f_o objective focal length (m), f_e eyepiece focal length, length L = f_o+f_e, final image at infinity, angular magnification ratio of angles subtended by image and object. For f_o=100 cm, f_e=5 cm, M=-20, inverted, used for celestial observation. Magnifying power: m = (f_o/f_e) . f_o = 150 cm , f_e = 5 cm . m = (150/5) = 30 . Substituting values gives 30, 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 > Telescope, Human Eye and Defects of Vision

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

A compound microscope has an objective of focal length \( 1.25 \, \text{cm} \) and eyepiece of focal length \( 5 \, \tex

**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) = (15/1.25) = 12 . Eyepiece magnification: m_e = (D/f_e) = (25/5) = 5 . Total magnification: m = m_o × m_e = 12 × 5 = 60 . Substituting values gives 60, 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 > Combination of Lenses and Lens Systems

A telescope has an objective of focal length \( 120 \, \text{cm} \) and an eyepiece of focal length \( 6 \, \text{cm} \)

**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. Magnifying power: m = (f_o/f_e) . f_o = 120 cm , f_e = 6 cm . m = (120/6) = 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), confirming sign conventions and refraction principles.

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

In a compound microscope, what role does the eyepiece play in the final image formation?

**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. The eyepiece in a compound microscope acts as a magnifying lens, taking the real, inverted image formed by the objective and producing a larger, virtual image for the observer. It enhances the angular size of the intermediate image, making it appear magnified without altering its orientation. Substituting values gives Magnifies the intermediate image, which matches expected image position and magnification

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

In a refracting telescope, why does the objective lens have a larger aperture than the eyepiece?

**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. The objective lens in a refracting telescope has a larger aperture to collect more light from distant objects, enhancing brightness and resolution. The eyepiece, with a smaller aperture, magnifies this image, requiring less light-gathering capacity for viewing. Substituting values gives To gather more light for brightness, which

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

A telescope has an objective of focal length \( 140 \, \text{cm} \) and an eyepiece of focal length \( 7 \, \text{cm} \)

**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. Magnifying power: m = (f_o/f_e) . f_o = 140 cm , f_e = 7 cm . m = (140/7) = 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), confirming sign conventions and refraction principles.

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

A compound microscope has an objective of focal length \( 1 \, \text{cm} \) and eyepiece of focal length \( 5 \, \text{c

**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. Objective magnification: m_o = (L/f_o) = (15/1) = 15 . Eyepiece magnification: m_e = (D/f_e) = (25/5) = 5 . Total magnification: m = m_o × m_e = 15 × 5 = 75 . Substituting values gives 75, 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

A telescope has an objective of focal length \( 100 \, \text{cm} \) and an eyepiece of focal length \( 5 \, \text{cm} \)

**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. Magnifying power: m = (f_o/f_e) . f_o = 100 cm , f_e = 5 cm . m = (100/5) = 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), confirming sign conventions and refraction principles.

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

A telescope has an objective of focal length \( 200 \, \text{cm} \) and an eyepiece of focal length \( 10 \, \text{cm} \

**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. Magnifying power: m = (f_o/f_e) . f_o = 200 cm , f_e = 10 cm . m = (200/10) = 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), confirming sign conventions and refraction principles.

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

A telescope has an objective of focal length \( 120 \, \text{cm} \) and an eyepiece of focal length \( 4 \, \text{cm} \)

**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. Magnifying power: m = (f_o/f_e) . f_o = 120 cm , f_e = 4 cm . m = (120/4) = 30 . Substituting values gives 30, 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

A telescope has an objective of focal length \( 180 \, \text{cm} \) and an eyepiece of focal length \( 6 \, \text{cm} \)

**Lens formula** 1/f = 1/v - 1/u, f focal length (m), u object distance, v image distance, sign: u negative if object left of lens in Cartesian convention, magnification m = v/u, real image inverted m negative, virtual erect m positive. Power P = 1/f (diopters D), f in meters, +5 D means f=0.2 m=20 cm converging. Magnifying power: m = (f_o/f_e) . f_o = 180 cm , f_e = 6 cm . m = (180/6) = 30 . Substituting values gives 30, 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

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