A thin convex lens of focal length 25cm is cut into two pieces 0.5cm above the principal axis. The top part is placed at (0, 0) and an object placed at (-50cm, 0). Find the coordinates of the image.
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Key concept: If a symmetric lens is cut parallel to principal axis in two parts. Focal length remains the same for each part. Intensity of image formed by each part will be less compared as that of complete lens. If there was no cut, then the object would have been at a height of 0.5cm from the principal axis OO'. 
The top part is placed at (0, 0) and an object placed at (-50cm, 0). There is no effect on the lenght of the lens.
Applying lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ $\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50}\Rightarrow\ \text{v}=50\text{cm}$ Magnification is $\text{m}=\frac{\text{v}}{\text{u}}=-\frac{50}{50}=-1$ Hence the image would have been formed at 50cm from the pole and 0.5cm below the principal axis. Hence, with respect to the X-axis passing through the edge of the cut lens, the coordinates of the image are (50cm, -1cm).
The top part is placed at (0, 0) and an object placed at (-50cm, 0). There is no effect on the lenght of the lens.Applying lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ $\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50}\Rightarrow\ \text{v}=50\text{cm}$ Magnification is $\text{m}=\frac{\text{v}}{\text{u}}=-\frac{50}{50}=-1$ Hence the image would have been formed at 50cm from the pole and 0.5cm below the principal axis. Hence, with respect to the X-axis passing through the edge of the cut lens, the coordinates of the image are (50cm, -1cm).
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