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 $Δ = 0.5 c m$ 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

Question

Verified by Toppr

Answer 1

For refraction from lens,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{- 1 0} = \frac{1}{2 0}$
$⟹ v = - 20 c m$ $= x_{1}, x_{2}$
$⟹ m = \frac{v}{u} = 2$
Now after displacement of the object, the lower end of the object is $0.5 c m$ below the optic axis,

Hence, $\frac{y _{1}}{- 0 . 5 c m} = 2$
$⟹ y_{1} = - 1 c m$
The upper end of the object lies $1 c m$ above the optic axis,
$\frac{y _{2}}{1 c m} = 2$
$⟹ y_{2} = 2 c m$

Answer 2

For refraction from lens,

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

\frac{1}{v} - \frac{1}{- 1 0} = \frac{1}{2 0}

⟹ v = - 20 c m

= x_{1}, x_{2}

⟹ m = \frac{v}{u} = 2

Now after displacement of the object, the lower end of the object is

0.5 c m

below the optic axis,

Hence,

\frac{y _{1}}{- 0 . 5 c m} = 2

⟹ y_{1} = - 1 c m

The upper end of the object lies

1 c m

above the optic axis,

\frac{y _{2}}{1 c m} = 2

⟹ y_{2} = 2 c m