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Q.The figure shows the image formation of an object in simple microscope.

(a) Find out the object distance and image distance from the figure.
(b) Derive an equation for magnifying power of the simple microscope. (1 + 2)
a simple microscope ray diagram forming a virtual magnified erect image with angles alpha and beta at the eye — Class 12 Physics question
Figure
Kerala DhseKerala DHSE Plus Two Board 2020Subjective· 3mImportance★★★★★
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In the figure, uu is the object's distance from the lens (just inside the focus) and vv is the virtual image's distance from the lens; the magnifying power works out to m=1+D/fm = 1+D/f when the image is formed at the near point.

  1. Object and image distances from the figure: The object PQPQ is placed between the lens centre OO and its focus FF, so the object distance is u=OQu = OQ (with u<fu < f). The refracted rays diverge after the lens and appear to come from the virtual, erect, magnified image P1Q1P_1Q_1 formed on the same side as the object; its distance from the lens is the image distance v=OQ1v = OQ_1 (with v>uv > u, and by lens sign convention both uu and vv are negative as measured from OO, but their magnitudes obey v>uv>u).
  2. Deriving the magnifying power: Angular magnification is defined as the ratio of the angle β\beta subtended by the image (through the lens) to the angle α\alpha subtended by the object when viewed directly at the near point DD (both angles measured at the eye, and for small angles tan⁡θ≈θ\tan\theta \approx \theta): m=βα=h/vh/D=Dv⋅vu⋅u1m = \frac{\beta}{\alpha} = \frac{h/v}{h/D} = \frac{D}{v}\cdot\frac{v}{u}\cdot\frac{u}{1} More directly, since β≈h/u\beta \approx h/u (angle subtended at the eye close to the lens by the image, essentially at the object's own transverse size hh over distance uu) — using similar triangles at the lens, m=v/um = v/u in general. …

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