Two objects A and B are placed at 15,cm and 2 | vedclass

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(C) Given: Radius of curvature $R = -40\,cm$ (for concave mirror). Focal length $f = R/2 = -20\,cm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{

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$160$

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(C) Given: Radius of curvature $R = -40\,cm$ (for concave mirror). Focal length $f = R/2 = -20\,cm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.For object $A$ at $u_1 = -15\,cm$:$\frac{1}{v_1} + \frac{1}{-15} = \frac{1}{-20} \implies \frac{1}{v_1} = -\frac{1}{20} + \frac{1}{15} = \frac{-3 + 4}{60} = \frac{1}{60}$.So,$v_1 = 60\,cm$ (image is virtual and behind the mirror).For object $B$ at $u_2 = -25\,cm$:$\frac{1}{v_2} + \frac{1}{-25} = \frac{1}{-20} \implies \frac{1}{v_2} = -\frac{1}{20} + \frac{1}{25} = \frac{-5 + 4}{100} = -\frac{1}{100}$.So,$v_2 = -100\,cm$ (image is real and in front of the mirror).The distance between the images is $d = |v_1 - v_2| = |60 - (-100)| = 160\,cm$.

Two objects $A$ and $B$ are placed at $15\,cm$ and $25\,cm$ from the pole in front of a concave mirror having a radius of curvature $40\,cm$. The distance between the images formed by the mirror is $..........\,cm$.

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