The objective lens of an optical instrument is an achromat combination with a focal length of $90 c m$ . The two lenses possess dispersive powers $0.024$ and $0.036$ respectively and are in contact with each other. Then their focal lengths are:

- 30 c m, 45 c m

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45 c m, 30 c m

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30 c m, - 45 c m

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30 c m, - 30 c m

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Solution

The correct option is B $30 c m, - 45 c m$
Given that:
$\frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{90}$ and $\frac{ω_{1}}{f_{1}} = - \frac{ω_{2}}{f_{2}}$

$\Rightarrow$ $\frac{0.024}{f_{1}} = - \frac{0.036}{f_{2}}$
$\Rightarrow f_{2} = - \frac{3}{2} f_{1}$

$\Rightarrow \frac{1}{f_{1}} + \frac{1}{- \frac{3}{2} f_{1}} = \frac{1}{90}$

$\Rightarrow \frac{1}{3 f_{1}} = \frac{1}{90}$

$\Rightarrow f_{1} = 30 c m$
$\Rightarrow f_{2} = - \frac{3}{2} \times 30 = - 45 c m$

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The objective lens of an optical instrument is an achromat combination with a focal length of 90 cm. The two lenses possess dispersive powers 0.024 and 0.036 respectively and are in contact with each other. Then their focal lengths are:

The correct option is B 30 cm,−45 cmGiven that: 1f1+1f2=190 and ω1f1=−ω2f2⇒0.024f1=−0.036f2⇒f2=−32f1⇒1f1+1−32f1=190 ⇒13f1=190⇒f1=30 cm⇒f2=−32×30=−45 cm

The objective lens of an optical instrument is an achromat combination with a focal length of $90 c m$ . The two lenses possess dispersive powers $0.024$ and $0.036$ respectively and are in contact with each other. Then their focal lengths are: