An object of height 4.0cm is placed at a distance of 30cm from the optical centre ‘O’ of a convex lens of focal length 20cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre ‘O’ and principal focus ‘F’ on the diagram. Also find the approximate ratio of size of the image to the size of the object.
CBSE DELHI - OUTSIDE DELHI - FOREIGN SET 3 2018
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Object size $\left(h_1\right)=4 cm$
Object distance (u) = -30cm
Focal length (f) = 20cm
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{+20}=\frac{1}{\text{v}}-\frac{1}{-30}$
$\frac{1}{\text{v}}=\frac{1}{20}-\frac{1}{30}=0.016$
⇒ Image distance (v)= 62cm
We know that,
$\text{m}=\frac{\text{height of image (h}_2)}{\text{height of object (h}_1)}=\frac{\text{v}}{\text{u}}=\frac{62}{-30}=-2.0$
Thus the approximate ratio of height of image to height of object is -2cm.
Negative sign denotes that image formed is inverted and real.
As the value of magnification is 2, the image is magnified.
Thus nature of image is real, inverted and magnified.
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