Question
Derive an expression for magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.
✓
Answer
Given: $T_A = 0.015 s, T_B = 0.025 s$
To find: greater frequency (n)
Formula: $n =\frac{1}{T}$
Calculation: From formula,
$ n _{ A }=\frac{1}{T_A}=\frac{1}{0.025}=\frac{1}{2.5 \times 10^{-2}}$
$\therefore n _{ A }=66.67 $
.... (using reciprocal table
$ n _{ B }=\frac{1}{T_B}=\frac{1}{0.025}=\frac{1}{2.5 \times 10^{-2}}$
$\therefore n _{ B }=40 Hz ....$ (using reciprocal table)
$\therefore n _{ A }> n _{ B }$
i. Figure (a) shows visual angle a made by an object, when kept at the least distance of distinct vision $(D = 25\ cm)$. Without an optical instrument this is the greatest possible visual angle as we cannot get the object closer than this.
ii. Figure (b) shows a convex lens forming erect, virtual and magnified image of the same object, when placed within the focus.
iii. The visual angle p of the object and the image in this case are the same. However, this time the viewer is looking at the image which is not closer than D. Hence the same object is now at a distance smaller than D.
iv. Angular magnification or magnifying power, in this case, is given by
$M =\frac{\text { Visualangleoftheimage }}{\text { Visualangleoftheobjectat } D}=\frac{\beta}{\alpha}$
For small angles,
$M =\frac{\beta}{\alpha} \approx \frac{\tan (\beta)}{\tan (\alpha)}=\frac{B A / P A}{B A / D}=\frac{D}{u}$
v. For maximum magnifying power, the image should be at D. For thin lens, considering thin lens formula.
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
In case of simple microscope,
$ f =+ f , v =- D , u =- u \text { and } M = M _{\max }$
$\therefore \quad \frac{1}{ f }=\frac{1}{- D }-\frac{1}{- u } $
As, $M=\frac{D}{ u }$
$M _{\max }=1+\frac{ D }{ f }$
vi. For minimum magnifying power, $v=\infty$ and $u = f$ (numerically)
$\therefore \quad M _{\min }=\frac{ D }{ u }=\frac{ D }{ f }$
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