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(B) $1$. For the convex lens: Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u = -50 \ cm$ and $f = +30 \ cm$. $\frac{1}{v} -

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$(50-25 \sqrt{3}, 25)$

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(B) $1$. For the convex lens: Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u = -50 \ cm$ and $f = +30 \ cm$. $\frac{1}{v} - \frac{1}{-50} = \frac{1}{30} \implies \frac{1}{v} = \frac{1}{30} - \frac{1}{50} = \frac{5-3}{150} = \frac{2}{150} = \frac{1}{75}$. Thus,$v = 75 \ cm$. $2$. This image acts as a virtual object for the convex mirror. The mirror is at $x = 50 \ cm$. The image formed by the lens is at $x = 75 \ cm$,which is $25 \ cm$ behind the mirror. So,for the mirror,the object distance $u = +25 \ cm$. $3$. For the convex mirror: $f_m = \frac{R}{2} = \frac{100}{2} = 50 \ cm$. Using the mirror formula $\frac{1}{v_m} + \frac{1}{u} = \frac{1}{f_m}$,where $u = +25 \ cm$ and $f_m = +50 \ cm$. $\frac{1}{v_m} + \frac{1}{25} = \frac{1}{50} \implies \frac{1}{v_m} = \frac{1}{50} - \frac{1}{25} = \frac{1-2}{50} = -\frac{1}{50}$. Thus,$v_m = -50 \ cm$. $4$. The image is formed $50 \ cm$ in front of the mirror along its tilted axis. The mirror's axis makes an angle $	heta = 30^{\circ}$ with the $x$-axis. The position of the mirror pole is $(50, 0)$. The image is at a distance of $50 \ cm$ from the pole along the axis at $30^{\circ}$. $5$. The coordinates of the image relative to the pole are: $\Delta x = -50 \cos 30^{\circ} = -50 	imes \frac{\sqrt{3}}{2} = -25\sqrt{3}$ and $\Delta y = 50 \sin 30^{\circ} = 50 	imes \frac{1}{2} = 25$. $6$. The absolute coordinates are $x = 50 - 25\sqrt{3}$ and $y = 25$. Thus,the coordinates are $(50 - 25\sqrt{3}, 25)$.

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