Practice question
Question
A concave mirror has a radius of curvature of \( 40 \, \text{cm} \). An object of height \( 2 \,
\text{cm} \) is placed \( 30 \, \text{cm} \) in front of it. What is the height of the image formed?
Explanation
**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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