Optical Instruments - Simple and Compound Microscope

Optical Instruments - Simple and Compound Microscope

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.

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Optical Instruments - Simple and Compound Microscope

An object of height \( 5 \, \text{cm} \) is placed \( 30 \, \text{cm} \) from a concave mirror of focal length \( 15 \,

**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. Focal length: f = -15 cm , u = -30 cm . Mirror equation: (1/v) + (1/-30) = (1/-15) ⇒ (1/v) = (1/-15) + (1/30) = (-2 + 1/30) = (-1/30) . v = -30 cm . Magnification: m = -(v/u) = -(-30/-30) = -1 . Image height: h' = m × h = -1 × 5 = -5 cm

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Optical Instruments - Simple and Compound Microscope

Why does the image in a refracting telescope appear inverted without additional optics?

**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. In a refracting telescope, the objective lens forms a real, inverted image of the distant object at its focal plane. The eyepiece magnifies this inverted image without reinverting it, so the final image remains inverted unless an additional lens or prism system is used. Substituting values gives

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Optical Instruments - Simple and Compound Microscope

An object of height \( 4 \, \text{cm} \) is placed \( 16 \, \text{cm} \) from a concave mirror of focal length \( 8 \, \

**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. Focal length: f = -8 cm , u = -16 cm . Mirror equation: (1/v) + (1/-16) = (1/-8) ⇒ (1/v) = (1/-8) + (1/16) = (-2 + 1/16) = (-1/16) . v = -16 cm . Magnification: m = -(v/u) = -(-16/-16) = -1 . Image height: h' = m × h = -1 × 4 = -4 cm (inverted). Magnitude =

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Optical Instruments - Simple and Compound Microscope

An object is placed \( 12 \, \text{cm} \) from a convex mirror of focal length \( 20 \, \text{cm} \). What is the magnif

**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. f = 20 cm , u = -12 cm . (1/v) + (1/-12) = (1/20) ⇒ (1/v) = (1/20) + (1/12) = (3 + 5/60) = (8/60) = (2/15) . v = 7.5 cm . Magnification: m = -(v/u) = -(7.5/-12) = 0.625 . Substituting values gives 0.625, which matches expected image position and magnification from mirror/lens formula 1/f =

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Optical Instruments - Simple and Compound Microscope

In a compound microscope, why is the final image inverted with respect to the object?

**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. In a compound microscope, the objective lens forms a real, inverted image of the object. The eyepiece then acts as a magnifying lens, forming a virtual image of this inverted intermediate image. Since the eyepiece does not reinvert the image, the final image remains inverted relative to the original object. Substituting values gives Due to the objective forming an inverted real image, which

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Optical Instruments - Simple and Compound Microscope

Why does the image in a reflecting telescope remain free of chromatic aberration?

**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. A reflecting telescope uses mirrors instead of lenses as the objective. Mirrors reflect all wavelengths of light uniformly, avoiding the wavelength-dependent refraction that causes chromatic aberration in lenses, resulting in a clearer, color-corrected image. Substituting values gives Due to uniform reflection of all wavelengths, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens)

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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

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Optical Instruments - Simple and Compound Microscope

An object of height \( 3 \, \text{cm} \) is placed \( 15 \, \text{cm} \) from a concave mirror of focal length \( 10 \,

**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. Focal length: f = -10 cm , u = -15 cm . Mirror equation: (1/v) + (1/-15) = (1/-10) ⇒ (1/v) = (1/-10) + (1/15) = (-3 + 2/30) = (-1/30) . v = -30 cm . Magnification: m = -(v/u) = -(-30/-15) = -2 . Image

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Optical Instruments - Simple and Compound Microscope

A concave mirror has a radius of curvature of \( 40 \, \text{cm} \). An object of height \( 2 \, \text{cm} \) is placed

**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. Focal length: f = (R/2) = (-40/2) = -20 cm (negative for concave mirror). Object distance: u = -30 cm . Using mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-30) = (1/-20) ⇒ (1/v) = (1/-20) + (1/30) = (-3 + 2/60) = (-1/60) . v = -60 cm (real image). Magnification: m = -(v/u) = -(-60/-30)

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Optical Instruments - Simple and Compound Microscope

A compound microscope has an objective of focal length \( 1 \, \text{cm} \) and eyepiece of focal length \( 4 \, \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) = (14/1) = 14 . Eyepiece magnification: m_e = (D/f_e) = (25/4) = 6.25 . Total magnification: m = m_o × m_e = 14 × 6.25 = 87.5 . Substituting values gives 87.5, 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.

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Optical Instruments - Simple and Compound Microscope

An object of height \( 3 \, \text{cm} \) is placed \( 20 \, \text{cm} \) from a concave mirror of focal length \( 15 \,

**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. f = -15 cm , u = -20 cm . (1/v) + (1/-20) = (1/-15) ⇒ (1/v) = (1/-15) + (1/20) = (-4 + 3/60) = (-1/60) . v = -60 cm . Magnification: m = -(v/u) = -(-60/-20) = -3 . Image height: h' = m × h = -3 × 3 = -9 cm (inverted). Magnitude = 9 cm . Substituting

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