- An object 3cm high is placed 24cm away from a convex lens of focal length 8cm. Find by calculations, the position, height and nature of the image.
- If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image?
- Which of the above two cases illustrates the working of a magnifying glass?
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- a. $h _1=3 cm$
f = 8cm
Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-24}=\frac{1}{8}$
$\frac{1}{\text{v}}=\frac{1}{12}$
$\text{v}=12\text{cm}$
Image is formed 12cm behind the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{12}{-24}=\frac{\text{h}_2}{3}$
$\text{h}_2=-1.5\text{cm}$
Image is 1.5cm hight, real and inverted.
- u = -3cm
f = 8cm
Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-3}=\frac{1}{8}$
$\frac{1}{\text{v}}=-\frac{5}{24}$
${\text{v}}=-4.8\text{cm}$
Image is formed 4.8cm in front of the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-4.8}{-3}=\frac{\text{h}_2}{3}$
$\text{h}_2=+4.8\text{cm}$
Image is 4.8cm high, virtual and erect.
- Case (b).
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