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(D) Let the focal length of the convex lens be $f_1 = 10 \, cm$ and the focal length of the concave lens be $f_2 = -f$ (where $f > 0$).According to th

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$-20$

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(D) Let the focal length of the convex lens be $f_1 = 10 \, cm$ and the focal length of the concave lens be $f_2 = -f$ (where $f > 0$).According to the lens combination formula,the effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.Given that the focal length of the combination is numerically equal to that of the concave lens,we have $F = -f$.Substituting the values: $\frac{1}{-f} = \frac{1}{10} + \frac{1}{-f}$.Wait,the problem states the combination's focal length is numerically equal to the concave lens. Let the concave lens focal length be $-x$. Then the combination focal length is $-x$.So,$\frac{1}{-x} = \frac{1}{10} - \frac{1}{x}$.Adding $\frac{1}{x}$ to both sides: $0 = \frac{1}{10}$. This implies a contradiction unless the combination focal length is defined differently.Re-evaluating: If the combination focal length $F$ is numerically equal to the concave lens focal length $f_c$,and $f_c$ is negative,then $F = f_c$.$\frac{1}{f_c} = \frac{1}{10} + \frac{1}{f_c}$. This is impossible.However,if the magnitude is equal,$|F| = |f_c|$. Since the combination is $F = \frac{10(-f)}{10-f}$,and we set $F = f$ (magnitude),then $f = \frac{10f}{f-10} \Rightarrow f-10 = 10 \Rightarrow f = 20$.Thus,the concave lens focal length is $-20 \, cm$.

$A$ convex lens of focal length $10 \, cm$ is placed in contact with a concave lens. The focal length of the combination is numerically equal to that of the concave lens. The focal length of the concave lens is ............ $cm$.

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