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?

Q: 1. 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: 1. 0.50m 2. 0.25m 3. 0.15m 2. Which of the above cases represents the use of convex lens in a film projector, in a camera, and as a magnifying glass?

1. f = 0.20m 1. u = 0.05m Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ $\frac{1}{\text{v}}-\frac{1}{-0.50}=\frac{1}{0.20}$ $\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.50}$ $\text{v}=0.33\text{m}$ 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. 2. $\text{u}=-0.25\text{m}$ Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ $\frac{1}{\text{v}}-\frac{1}{-0.25}=\frac{1}{0.20}$ $\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.25}$ $\text{v}=1\text{m}$ Image is formed 1m behind the lens. $\text{m}=\frac{\text{v}}{\text{u}}=\frac{1}{-0.25}=-4$ Image is real and inverted. 3. $\text{u}=-0.15\text{m}$ Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ $\frac{1}{\text{v}}-\frac{1}{-0.15}=\frac{1}{0.20}$ $\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.15}$ $\text{v}=-0.60\text{m}$ 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. 2. Film projector, Case (ii) Camera, Case (i) Magnifying glass, Case (iii)

Question

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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Solution

f = 0.20m

u = 0.05m

Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{v}}-\frac{1}{-0.50}=\frac{1}{0.20}$

$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.50}$

$\text{v}=0.33\text{m}$

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.

$\text{u}=-0.25\text{m}$

Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{v}}-\frac{1}{-0.25}=\frac{1}{0.20}$

$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.25}$

$\text{v}=1\text{m}$

Image is formed 1m behind the lens.

$\text{m}=\frac{\text{v}}{\text{u}}=\frac{1}{-0.25}=-4$

Image is real and inverted.

$\text{u}=-0.15\text{m}$

Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{v}}-\frac{1}{-0.15}=\frac{1}{0.20}$

$\frac{1}{\text{v}}=\frac{1}{0.20}-\frac{1}{0.15}$

$\text{v}=-0.60\text{m}$

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)

Camera, Case (i)

Magnifying glass, Case (iii)

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