An object is placed at a distance of 30 cm from a concave lens of focal length 15 cm. List four characteristics (nature, position, etc.) of the image formed by the lens.
CBSE OUTSIDE DELHI - SET 1 2017
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U = -30cm f = -15cm $\frac{1}{\text{f}} = \frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}} = \frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\Big(\frac{1}{-15}\Big)+\Big(\frac{1}{-30}\Big)$
$\frac{1}{\text{v}}=\Big(\frac{1}{-10}\Big)$
On substituting v = -10cm$\text{m}=\frac{\text{v}}{\text{u}}$
$\text{m}=\frac{-10}{-30}$
$\text{m}=\frac{1}{3}=.33$
The magnification is positive so it is a virtual erect image, m is less than 1 so the image is diminished and v is negative so the image is formed on the same side of the lens.
$\frac{1}{\text{v}} = \frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\Big(\frac{1}{-15}\Big)+\Big(\frac{1}{-30}\Big)$
$\frac{1}{\text{v}}=\Big(\frac{1}{-10}\Big)$
On substituting v = -10cm$\text{m}=\frac{\text{v}}{\text{u}}$
$\text{m}=\frac{-10}{-30}$
$\text{m}=\frac{1}{3}=.33$
The magnification is positive so it is a virtual erect image, m is less than 1 so the image is diminished and v is negative so the image is formed on the same side of the lens.
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