Derive the formula for angular magnification of a compound microscope, when the final image is formed at least distance of distinct vision. Draw the required ray diagram.
The angular magnification of a compound microscope is the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye, when both are placed at the least distance of distinct vision. 
Focal length of the objective lens, f1 = 2.0 cm Focal length of the eyepiece, f2 = 6.25 cm Distance between the objective lens and the eyepiece, d = 15 cm (a) Least distance of distinct vision, d' = 25 ∴ Image distance for the eyepiece, v2 = −25 cm Object distance for the eyepiece = u2 According to the lens formula, we have the relation: `1/"v"_2 - 1/"u"_2 = 1/"f"_2` `1/"u"_2 = 1/"v"_2 - 1/"f"_2` = `1/(-25) - 1/6.25` = `(-1 - 14)/25` = `(-5)/25` ∴ u2 = −5 cm Image distance for the objective lens, v1 = d + u2 = 15 − 5 = 10 cm Object distance for the objective lens = u1 According to the lens formula, we have the relation: `1/"v"_1 - 1/"u"_1 = 1/"f"_1` `1/"u"_1 = 1/"v"_1 - 1/"f"_1` = `1/10 - 1/2` = `(1 - 5)/10` = `(-4)/10` ∴ u1 = −2.5 cm Magnitude of the object distance, |u1| = 2.5 cm The magnifying power of a compound microscope is given by the relation: `"m" = "v"_1/|"u"_1| (1 + "d'"/"f"^2)` = `10/2.5 (1+ 25/6.25)` = 4 × (1 + 4) = 20 Hence, the magnifying power of the microscope is 20. (b) The final image is formed at infinity. ∴ Image distance for the eyepiece, v2 = ∞ Object distance for the eyepiece = u2 According to the lens formula, we have the relation: `1/"v"_2 - 1/"u"_2 = 1/"f"_2` `1/∞ - 1/"u"_2 = 1/6.25` ∴ u2 = −6.25 cm Image distance for the objective lens, v1 = d + u2 = 15 − 6.25 = 8.75 cm Object distance for the objective lens = u1 According to the lens formula, we have the relation: `1/"v"_1 - 1/"u"_1 = 1/"f"_1` `1/"u"_1 = 1/"v"_1 - 1/"f"_1` = `1/8.75 - 1/2.0` = `(2 - 8.75)/17.5` ∴ `"u"_1 = -17.5/6.75` = −2.59 cm Magnitude of the object distance, |u1| = 2.59 cm The magnifying power of a compound microscope is given by the relation: `"m" = "v"_1/|"u"_1| (("d'")/|"u"_2|)` = `8.75/2.59 xx 25/6.25` = 13.51 Hence, the magnifying power of the microscope is 13.51.
What is the magnifying power of a simple microscope of focal length $5 \mathrm{~cm}$, if the image is formed at the distance of distinct vision? A)4 B)7 C)6 D)5
Correct option is C Explanation: Focal length of convex lens, $f=5 \mathrm{~cm}$ Position of image, $\mathrm{v}=25 \mathrm{~cm}$ Position of object, $\mathrm{u}=$ ? Using lens formula $ \begin{array}{l} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\ \frac{1}{5}=\frac{1}{25}-\frac{1}{\mathrm{u}} \\ \mathrm{u}=\frac{25}{4} \mathrm{~cm} \end{array} $ Magnifying power of instrument: $ \begin{array}{l} \mathrm{M}=1+\frac{\mathrm{D}}{\mathrm{f}} \\ \mathrm{M}=1+\frac{25}{5}=6 \end{array} $ Hence, object distance is $\frac{25}{4} \mathrm{~cm}$ and magnifying power is 6 |
What is the magnifying power of simple microscope of focal length 5cm when image is formed at the distance of distinct vision a 7 B 5 C 6 D 4?
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