Question

The image of a point object kept on the principal axis of a concave mirror of radius of curvature $R=40cm$ is real, inverted and magnified. If the difference in distance between image and object is 30 cm, as per sign convention the object distance (u) will be:

• A. $u=30cm$
• B. $u=-30cm$
• C. $u=60cm$
• D. $u=-60cm$

Answer 1

The correct option is B $u=-30cm$

Given,
Radius of curvature $R=40cm$
As,
f = $\frac{R}{2}$

Focal length of the mirror (f) = $\frac{R}{2}$
$\Rightarrow \frac{40}{2}=20cm$.

Let u be the distance of image from the pole of the concave mirror. As it is given that distance between the image and the object is 30 cm and a real, inverted and magnified image is being formed by the concave mirror, so the object must have been kept between centre of curvature and focus of the mirror. So as per sign convention
i.e, $v=-(u+30)$

Using mirror formula,
$\frac{1}{\text{f}}=$ $\frac{1}{\text{v}}+$ $\frac{1}{\text{u}}$
Now using sign convention,
$\frac{1}{-20}=$ $\frac{1}{-(u+30)}+$ $\frac{1}{\text{-u}}$
$=u^2-10u-600=0$
solving the equation we will get,
$u=-20cm$ and $u=30cm$

It is important to note here that we considered u as the distance of the object from the pole. So, u = 30 cm is the correct solution as u = - 20 is not possible.
Thus object distance from the pole of the mirror is 30 cm or u = -30 cm.