Question

A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size and nature of the image.

• A. $\frac{10}{3}cm$ from the mirror on the side opposite to the object, virtual and erect,
• B. $\frac{10}{3}cm$ fro the mirror on the same side as the object, virtual and erect.
  
• C. $\frac{10}{3}cm$ from the mirror on the side opposite to the object, virtual and inverted.
• D. $\frac{10}{3}cm$ from the mirror on the side opposite to the object, real and inverted.

Answer 1

The correct option is A $\frac{10}{3}cm$ from the mirror on the side opposite to the object, virtual and erect,

Let's Draw the schematic first.

First of all to find the location of the image we can apply mirror formula.
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
Or
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
So,
$\frac{1}{v}=\frac{1}{6}-\frac{-1}{7.5}$
On solving we get,
$\frac{1}{v}=\frac{1}{6}+\frac{1}{7.5}$
$\frac{1}{v}=\frac{3}{10}c{m}^{-1}$
Or,
$v=\frac{10}{3}cm$.
Now the positive value of v means, the image is behind the mirror according to our sign convention.
Now let's find magnification and see what we get from there,
$m=\frac{h_i}{h_0}=-\frac{v}{u}$
$\frac{h_i}{h_0}=\frac{\frac{-10}{3}}{-7.5}=\frac{10}{22.5}$
We see that the height of the image is positive which means that the image is erect and hence it has to be virtual.