An equiconvex lens of focal length 10 cm is cut into two equal halves by cutting along a plane perpendicular to the principal axis. The two pieces are put 10 cm apart with curved surfaces facing each other. The focal length of the new combination is

8 c m

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20 c m

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\frac{40}{3} c m

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\frac{10}{2} c m

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Solution

The correct option is D $\frac{40}{3} c m$
$f e q = \frac{f_{1} f_{2}}{f_{1} + f_{2} - d}$

when a lens is cut f doubles.

$f e q = \frac{20 \times 20}{20 + 20 - 10} = \frac{400}{30} = \frac{40}{3} c m$

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An equiconvex lens of focal length 10 cm is cut into two equal halves by cutting along a plane perpendicular to the principal axis. The two pieces are put 10 cm apart with curved surfaces facing each other. The focal length of the new combination is

The correct option is D 403cmfeq=f1f2f1+f2−dwhen a lens is cut f doubles.feq=20×2020+20−10=40030=403cm

The correct option is D $\frac{40}{3} c m$
$f e q = \frac{f_{1} f_{2}}{f_{1} + f_{2} - d}$

when a lens is cut f doubles.

$f e q = \frac{20 \times 20}{20 + 20 - 10} = \frac{400}{30} = \frac{40}{3} c m$