An object is placed 30 cm away from a convex | vedclass

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(D) For the first condition, using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:Given $f_1 = 10 \ cm$ and $u = -30 \ cm$.$\frac{1}{10} =

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) For the first condition, using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:Given $f_1 = 10 \ cm$ and $u = -30 \ cm$.$\frac{1}{10} = \frac{1}{v} - \frac{1}{-30} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$.So, $v = 15 \ cm$.For the second condition, when the concave lens is placed in contact, the new image distance $v' = v + 45 \ cm = 15 + 45 = 60 \ cm$.The object distance $u$ remains $-30 \ cm$.Let $F$ be the focal length of the combination: $\frac{1}{F} = \frac{1}{v'} - \frac{1}{u} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}$.So, $F = 20 \ cm$.Using the combination formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$:$\frac{1}{20} = \frac{1}{10} + \frac{1}{f_2} \Rightarrow \frac{1}{f_2} = \frac{1}{20} - \frac{1}{10} = \frac{1-2}{20} = -\frac{1}{20}$.Thus, $f_2 = -20 \ cm$.The magnitude of the focal length of the concave lens is $20 \ cm$.

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