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An object is placed at a distance of $30\,\text{cm}$ in front of a convex lens of focal length $15\,\text{cm}$. Use lens formula to determine the position of the image. What will be the size of the image in this case ?
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• $u = -30\text{ cm}$, $f = 15\text{ cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{15} + \frac{1}{-30}$ (½ Mark)
$v = 30\text{ cm}$ (1 Mark)
• $m = \frac{v}{u} = +\frac{30}{-30} = -1$
Hence, size of image will be same as the object. / Size of image will be same as size of object because when object is kept at $2F_1$, image will be formed at $2F_2$. (1 Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{15} + \frac{1}{-30}$ (½ Mark)
$v = 30\text{ cm}$ (1 Mark)
• $m = \frac{v}{u} = +\frac{30}{-30} = -1$
Hence, size of image will be same as the object. / Size of image will be same as size of object because when object is kept at $2F_1$, image will be formed at $2F_2$. (1 Mark)