An object is placed at a distance of 40, cm i | vedclass

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(A) Given: Object distance $u = -40\, cm$,Focal length $f = -20\, cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting

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real,inverted and of the same size as that of the object

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(A) Given: Object distance $u = -40\, cm$,Focal length $f = -20\, cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the values: $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-40}$.$\frac{1}{v} = \frac{1}{40} - \frac{1}{20} = \frac{1 - 2}{40} = -\frac{1}{40}$.So,$v = -40\, cm$.The magnification $m = -\frac{v}{u} = -\frac{-40}{-40} = -1$.$A$ magnification of $-1$ indicates that the image is real,inverted,and of the same size as the object.

An object is placed at a distance of $40\, cm$ in front of a concave mirror of focal length $20\, cm$. The image produced is

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