An angular magnification (magnifying power) of 30 x is desired for a c

1.	An angular magnification (magnifying power) of 30 x is desired for a compound microscope using as objective of focal length 1.25 cm and eye piece of focal length 5 cm. How will you set up the compound microscope?
Answer» ∵ m = m_o m_e = -30 (virtual, inverted) ∵ f_o = 1.25cm f_e = 5.0cm Let us setup a compound microscope such that the final image be formed at D, then m_e = 1 + (D/f_e) = 1 + (25/5) = 6 and position of object for this image formation can be calculated – (i/Ve) - (1/ue) = 1/f_e (1/-25) - (1/ue) = 1/5 - 1/ue = (1/5) + (1/25) = 6/25 ue = -25/6 = - 4.17 cm ∵ m = m_o × m_e ∴ m_o = +Vo/uo = -30/6 = -5 ∴ V = -5u_o (1/Vo) - (1/uo) = 1/fo (1/-5uo) - (1/uo) = 1/1.25 -6/5uo = 1/1.25 uo = -1.5cm ⟹ v_o= 7.5cm Tube length = V_o + \|u_e\| = 7.5 cm + 4.17 cm L = 11.67 cm Object should be placed at 1.5 cm distance from the objective lens.

An angular magnification (magnifying power) of 30 x is desired for a compound microscope using as objective of focal length 1.25 cm and eye piece of focal length 5 cm. How will you set up the compound microscope?

Answer»

∵ m = m_o m_e = -30 (virtual, inverted)

∵ f_o = 1.25cm

f_e = 5.0cm

Let us setup a compound microscope such that the final image be formed at D, then

m_e = 1 + (D/f_e) = 1 + (25/5) = 6

and position of object for this image formation can be calculated –

(i/Ve) - (1/ue) = 1/f_e

(1/-25) - (1/ue) = 1/5

- 1/ue = (1/5) + (1/25) = 6/25

ue = -25/6 = - 4.17 cm

∵ m = m_o × m_e

∴ m_o = +Vo/uo = -30/6 = -5

∴ V = -5u_o

(1/Vo) - (1/uo) = 1/fo

(1/-5uo) - (1/uo) = 1/1.25

-6/5uo = 1/1.25

uo = -1.5cm ⟹ v_o= 7.5cm

Tube length = V_o + |u_e| = 7.5 cm + 4.17 cm

L = 11.67 cm

Object should be placed at 1.5 cm distance from the objective lens.

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