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(B) Given: Object distance $u = -50\, cm$,Image distance $v = -10\, cm$ (since the virtual image is formed in front of the lens).Using the lens formul

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$-12.5$

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(B) Given: Object distance $u = -50\, cm$,Image distance $v = -10\, cm$ (since the virtual image is formed in front of the lens).Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-10} - \frac{1}{-50}$.$\frac{1}{f} = -\frac{1}{10} + \frac{1}{50} = \frac{-5 + 1}{50} = \frac{-4}{50}$.Therefore,$f = -\frac{50}{4} = -12.5\, cm$.The negative sign indicates that the lens is a concave lens.

An object is placed $50\, cm$ from a lens that produces a virtual image at a distance of $10\, cm$ in front of the lens. Calculate the focal length of the lens. (in $, cm$)

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