A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula.
CBSE OUTSIDE DELHI - SET 1 2012
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Given: Height of object, h = 4cm Focal length of convex lens, f = +24cm Object distance, u = -16cm Applying the lens formula: $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\frac{1}{\text{v}}-\frac{1}{-16}=\frac{1}{24}$
On calculating we get, $\frac{1}{\text{v}}=-\frac{1}{48}$$\text{V}=-48\text{cm}$
Now size of the image can be calculated by: Magnification is given by: $\frac{\text{v}}{\text{u}}=\text{m}=\frac{\text{h}_2}{\text{h}_1}$$\frac{\text{h}_2}{4}=-\frac{48}{-16}$
On calculating $h _2=12 cm$ Therefore, the image will be fornme as virtual, erect and magnified.
On calculating we get, $\frac{1}{\text{v}}=-\frac{1}{48}$$\text{V}=-48\text{cm}$
Now size of the image can be calculated by: Magnification is given by: $\frac{\text{v}}{\text{u}}=\text{m}=\frac{\text{h}_2}{\text{h}_1}$$\frac{\text{h}_2}{4}=-\frac{48}{-16}$
On calculating $h _2=12 cm$ Therefore, the image will be fornme as virtual, erect and magnified.
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