Question

Choose the correct answer from alternatives given.
A linear object of size 1.5 cm is placed at 10 cm from a lens of focal length 20 cm. The optic centre of lens and the object are displaced a distance $\Delta$ The magnification of the image formed is m. (Take optic centre as origin). The coordinates of image of A and B are $(x_1,y_1)$ and $(x_2,y_2)$ spectively.Then

• A. $(x_1,y_1)=(20cm,-2cm)$
• B. $(x_2,y_2)=(-20cm,2cm)$
• C. m = 5
• D. m = 4

Answer 1

The correct option is B $(x_2,y_2)=(-20cm,2cm)$

u = -10 cm, f = 20 cm

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f};\frac{1}{v}=\frac{1}{20}-\frac{1}{10}=\frac{-1}{20}\Rightarrow v=-20cm$
Magnification, $m=\frac{v}{u}=\frac{-20}{-10}=2$
Height of image = 1.5 \times 2 = 3 cm
The $y_1$ coordinate of a point A on the image will be $y_1=(-0.5)\times 2=-1cm$
The $y_2$ coordinate of a point B on the image will be
$y_2=1\times 2=2cm$