Question

A linear object of size 1.5 cm placed at 10 cm from a lens of focal length 20 cm. The optic centre of lens and the object are displaced a distance $\Delta =0.5cm$ as shown in the figure. 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 $(x_2,y_2)$ respectively. Then

• A. $m=3$
• B. $(x_1,y_1)=(-20cm,-1cm)$
• C. $(x_2,y_2)=(-20cm,2cm)$
• D. $m=2$

Answer 1

Answer: (B), (C), (D)

For refraction from lens,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{-10}=\frac{1}{20}$

$⟹v=-20cm$ $=x_1,x_2$

$⟹m=\frac{v}{u}=2$

Now after displacement of the object, the lower end of the object is $0.5cm$ below the optic axis,

Hence, $\frac{y_1}{-0.5cm}=2$

$⟹y_1=-1cm$

The upper end of the object lies $1cm$ above the optic axis,

$\frac{y_2}{1cm}=2$

$⟹y_2=2cm$