Question

find the distance between two position of a real object, for which a convex lens of power 2.5 D produces an image that is four times as large as the object

Answer 1

Answer: 150 cm

For a thin converging lens,

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

where, f is the focal length of the lens, u is the object distance from lens and v is the image distance from the lens.

Magnification, $M=\frac{h_i}{h_o}=-\frac{v}{u}$

where,  $h_i, h_o$ is the height of the image and object respectively.

Power, $P(D)=\frac{1}{f(m)}$

We need to find the distance between the object and its image, i.e. (u-v).

It is given that, $P=\frac{1}{f(m)}=2.5 D$

and $h_i=4h_o \Rightarrow M=-\frac{v}{u}=\frac{4h_o}{h_o}=4\Rightarrow v=-4u$

Using the given information, substitute in the lens equation,

$\Rightarrow \frac{1}{-4u}+\frac{1}{u}=2.5 \\ \Rightarrow 2.5u=1-\frac{1}{4}\Rightarrow u=0.75\times 0.4=.3 m=30 cm$

$\Rightarrow v=-4u=-120 cm$

Therefore, $u-v=30-(-120)cm=150 cm=1.5 m$