The focal length of a concave mirror is 30 , | vedclass

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(A) $(i)$ If the image is real and inverted:$m = -v/u = -3$,which implies $v = 3u$. Given $f = -30 \, cm$.Using the mirror formula $\frac{1}{v} + \fra

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$40 \, cm$ and $20 \, cm$

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(A) $(i)$ If the image is real and inverted:$m = -v/u = -3$,which implies $v = 3u$. Given $f = -30 \, cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{3u} + \frac{1}{u} = \frac{1}{-30} \Rightarrow \frac{1+3}{3u} = \frac{1}{-30} \Rightarrow \frac{4}{3u} = \frac{1}{-30} \Rightarrow u = -40 \, cm$.Thus,the object is placed at $40 \, cm$ in front of the mirror.$(ii)$ If the image is virtual and erect:$m = -v/u = +3$,which implies $v = -3u$. Given $f = -30 \, cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{-3u} + \frac{1}{u} = \frac{1}{-30} \Rightarrow \frac{-1+3}{3u} = \frac{1}{-30} \Rightarrow \frac{2}{3u} = \frac{1}{-30} \Rightarrow u = -20 \, cm$.Thus,the object is placed at $20 \, cm$ in front of the mirror.

The focal length of a concave mirror is $30 \, cm$. Find the position of the object in front of the mirror so that the image is three times the size of the object.

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