A 2.0cm tall object is place perpendicular to the principal axis of a concave lens of focal length 15cm. At what distance from the lens, should the object be placed so that it forms an image 10cm from the lens? Also find the nature and the size of image formed.
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For the object, h = 2cm, f = 10cm, u = -15cm From lens formula,$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\therefore\ \frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{10}+\frac{1}{-15}=\frac{1}{10}-\frac{1}{15}$
$\therefore\ \frac{1}{\text{v}}=\frac{3-2}{30}=\frac{1}{30}$
$\therefore\ \text{v}=30\text{cm}$
As v is positive, the image is real. The other aspects about the nature canbe determined from the value of magnification.$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}'}{\text{h}}$
$\therefore\ \frac{30}{-15}=\frac{\text{h}'}{2}$
$\therefore\ \text{h}'=-4$
Hence, the height of the image is 4cm. This suggests that the image is enlarged. The negative sign of the magnification suggests that the image is inverted. Hence, the image is real, inverted and enlarged.
$\therefore\ \frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{10}+\frac{1}{-15}=\frac{1}{10}-\frac{1}{15}$
$\therefore\ \frac{1}{\text{v}}=\frac{3-2}{30}=\frac{1}{30}$
$\therefore\ \text{v}=30\text{cm}$
As v is positive, the image is real. The other aspects about the nature canbe determined from the value of magnification.$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}'}{\text{h}}$
$\therefore\ \frac{30}{-15}=\frac{\text{h}'}{2}$
$\therefore\ \text{h}'=-4$
Hence, the height of the image is 4cm. This suggests that the image is enlarged. The negative sign of the magnification suggests that the image is inverted. Hence, the image is real, inverted and enlarged.
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