Name: Chromatic Aberration
Uploaded: 2019-07-19
Duration: 5 min 45 s
Description: Do you know what chromatic aberration in lenses is? How can we reduce this? Let's dive into the concept behind it and learn to avoid it in the future.

Question

Focal power of the eye lens of a compound microscope is

6

dioptre. The microscope is to be used for maximum magnifying power (angular magnification) of at least

12.5

. The packing instructions demand that length of the microscope should be

25

cm. Determine minimum focal power of the objective. How much will its radius of curvature be if it is a biconvex lens of

n = 1.5

.

Accepted Answer

Given that $P_{eye piece} = 6 D$

$∴ f = \frac{1}{P} = \frac{1}{6} m = \frac{100}{6} cm$

$L =$ length of microscope $= 25 cm$

For max power $V_{e} = D = 25 cm$ .

Lens formula for eyelens :

$\frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}}$

Using sign convention

$- \frac{1}{u_{e}} = \frac{1}{v_{e}} - \frac{1}{f_{e}} ∴ - \frac{1}{u_{e}} = \frac{1}{- 25} - \frac{6}{100} ∴ - \frac{1}{u_{e}} = \frac{- 4 - 6}{100} = - \frac{10}{100} ∴ \frac{1}{u_{e}} = \frac{1}{10} ∴ u_{e} = 10 cm$

Now

$L = v_{0} + u_{e} ∴ 25 = v_{0} + 10 ∴ v_{0} = 15 cm$

Max. magnifying power

$= \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}})$

$∴ 12.5 = \frac{15}{u_{0}} (1 + \frac{25 \times 6}{100}) ∴ \frac{12.5}{15} = \frac{2.5}{u_{0}} ∴ u_{0} = \frac{2.5 \times 15}{12.5} ∴ u_{0} = 3 cm$

Now

$\frac{1}{f_{0}} = \frac{1}{v_{0}} - \frac{1}{u_{0}} ∴ \frac{1}{f_{0}} = \frac{1}{15} - (\frac{1}{- 3}) ∴ \frac{1}{f_{0}} = \frac{1}{15} + \frac{1}{3} ∴ \frac{1}{f_{0}} = \frac{1 + 5}{15} ∴ \frac{1}{f_{0}} = \frac{6}{15} ∴ \frac{1}{f_{0}} = \frac{2}{5} ∴ f_{0} = \frac{5}{2} cm = 2.5 cm ∴ f_{0} = \frac{5}{2} \times 10^{- 2} m ∴ f_{0} = \frac{5}{200} m ∴ f_{0} = \frac{1}{40} m$

Now $P_{0} = \frac{1}{f_{0}} = 40 D$

If $n = 1.5$ ,

Using lens maker's formula,

$\frac{1}{f} = (n - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (∵ R_{1} = - R_{2} = R) ∴ \frac{1}{f} = (1.5 - 1) (\frac{1}{R} + \frac{1}{R}) ∴ \frac{1}{f} = (0.5) (\frac{2}{R}) ∴ f = \frac{R}{2 \times 0.5} ∴ f = R ∴ R = 2.5 cm$

Hence, Radius of curvature of biconvex lens is $R = 2.5 cm$ .

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