The distance between an object and its two ti | vedclass

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(A) For a real image,magnification $m = -2$.Since $m = \frac{v}{u}$,we have $\frac{v}{u} = -2$,which implies $v = -2u$.Given the distance between the

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$10$

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(A) For a real image,magnification $m = -2$.Since $m = \frac{v}{u}$,we have $\frac{v}{u} = -2$,which implies $v = -2u$.Given the distance between the object and the image is $45 \,cm$,and for a real image formed by a convex lens,the object and image are on opposite sides,the distance is $|v| + |u| = 45$.Substituting $v = -2u$ (where $u$ is negative),we get $|-2u| + |u| = 45$,so $3|u| = 45$,which gives $|u| = 15 \,cm$.Thus,$u = -15 \,cm$ and $v = +30 \,cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{30} - \frac{1}{-15} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$.Therefore,$f = 10 \,cm$.

The distance between an object and its two times magnified real image produced by a convex lens is $45 \,cm$. The focal length of the lens used is . . . . . . $cm$.

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