- Find the nature, position and magnification of the images formed by a convex lens of focal length 0.20 m if the object is placed at a distance of:
- 0.50m
- 0.25m
- 0.15m
- Which of the above cases represents the use of convex lens in a film projector, in a camera, and as a magnifying glass?
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- f = 0.20m
- u = 0.05m
$\frac{1}{\text{v}}-\frac{1}{-0.50}=\frac{1}{0.20}$
$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.50}$
$\text{v}=0.33\text{m}$
Image is formed 0.33m behind the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{0.33}{-0.50}=-0.66$
Image is real and inverted.
- $\text{u}=-0.25\text{m}$
$\frac{1}{\text{v}}-\frac{1}{-0.25}=\frac{1}{0.20}$
$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.25}$
$\text{v}=1\text{m}$
Image is formed 1m behind the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{1}{-0.25}=-4$
Image is real and inverted.
- $\text{u}=-0.15\text{m}$
$\frac{1}{\text{v}}-\frac{1}{-0.15}=\frac{1}{0.20}$
$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.15}$
$\text{v}=-0.60\text{m}$
Image is formula 0.60m in front of the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{-0.6}{-0.15}=+4$
Image is virtual and erect.
- Film projector, Case (ii)
Magnifying glass, Case (iii)
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