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

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Question

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

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Real,inverted,and of the same size

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

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

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Assertion : The formula connecting $u, v$ and $f$ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.
Reason : Laws of reflection are strictly valid for plane surfaces,but not for large spherical surfaces.

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