Question

A convex lens forms an image of an object placed $20\text{cm}$ away from it at a distance of $20\text{cm}$ on the other side of the lens. If the object is moved $5\text{cm}$ towards the lens, the image will move

• A. $5\text{cm}$ toward the lens
• B. $5\text{cm}$ away from the lens
• C. $10\text{cm}$ towards the lens
• D. $10\text{cm}$ away from the lens

Answer 1

The correct option is D $10\text{cm}$ away from the lens

From the given problem,
Object distance, $u=-20\text{cm}$
Image distance, $v=+20\text{cm}$

Applying the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Substituting the values of $u$ and $v$, we have
$\frac{1}{20}-\frac{1}{-20}=\frac{1}{f}$

$\therefore f=10\text{cm}$

Now, when object is moved $5\text{cm}$ towards lens,
then, object distance, $u_1=-(20-5)\text{cm}=-15\text{cm}$
Let, the image distance is $v_1$
Applying lens formula,
$\frac{1}{v_1}-\frac{1}{-15}=\frac{1}{10}$

$\Rightarrow \frac{1}{v_1}+\frac{1}{15}=\frac{1}{10}$

$\Rightarrow \frac{1}{v_1}=\frac{1}{10}-\frac{1}{15}$

$\Rightarrow \frac{1}{v_1}=\frac{3-2}{30}=\frac{1}{30}$

$\therefore v_1=30\text{cm}$

The image will move by distance $(30-20)\text{cm}$ $i.e.$ $10\text{cm}$ away from the lens.