A converging lens of focal length $40$ cm is kept in contact with a diverging lens of focal length $30$ cm. Find the focal length of the combination.

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Solution

Focal length of the converging lens is $40$ cm i.e. $f_{1} = 40$ cm

Focal length of the diverging lens is $30$ cm i.e. $f_{2} = - 30$ cm

Thus focal length of combination $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

$∴$ $\frac{1}{F} = \frac{1}{40} + \frac{1}{(- 30)}$

$⟹$ $F = - 120$ cm

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A converging lens of focal length 40cm is kept in contact with a diverging lens of focal length 30cm. Find the focal length of the combination.

Focal length of the converging lens is 40 cm i.e. f1=40 cmFocal length of the diverging lens is 30 cm i.e. f2=−30 cmThus focal length of combination 1F=1f1+1f2∴ 1F=140+1(−30)⟹ F=−120 cm

A converging lens of focal length $40$ cm is kept in contact with a diverging lens of focal length $30$ cm. Find the focal length of the combination.

Focal length of the converging lens is $40$ cm i.e. $f_{1} = 40$ cm

Focal length of the diverging lens is $30$ cm i.e. $f_{2} = - 30$ cm

Thus focal length of combination $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

$∴$ $\frac{1}{F} = \frac{1}{40} + \frac{1}{(- 30)}$

$⟹$ $F = - 120$ cm