Question

A $3\text{cm}$ long object is placed at $12\text{cm}$ from a concave lens, perpendicular to its principal axis. The lens forms a virtual image whose size is $1.5\text{cm}$. Find the power of the lens.

• A. $-\frac{25}{3}\text{D}$
• B. $-8\text{D}$
• C. $-\frac{7}{36}\text{D}$
• D. $-\frac{25}{9}\text{D}$

Answer 1

The correct option is A $-\frac{25}{3}\text{D}$

Given:

$u=-12\text{cm}$
$h_0=3\text{cm}$

Since the lens forms a virtual image, so the image is erect.

$\therefore h_i=1.5\text{cm}$

Now,

$m=\frac{v}{u}=\frac{h_i}{h_0}$

$\Rightarrow \frac{v}{-12}=\frac{1.5}{3}$

$\Rightarrow v=-6\text{cm}$

Using lens formula,

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

$\Rightarrow \frac{1}{(-6)}-\frac{1}{(-12)}=\frac{1}{f}$

$\Rightarrow \frac{1}{f}=\frac{1}{12}-\frac{1}{6}=\frac{-1}{12}$

$\Rightarrow f=-12\text{cm}=-\frac{12}{100}\text{m}$

Now, power,

$P=\frac{1}{f}$
$\Rightarrow P=-\frac{100}{12}=-\frac{25}{3}\text{D}$