Question

A $2.0cm$ tall object is placed $15cm$ in front of a concave mirror of focal length $10cm$. What is the size and nature of the image?

• A. 4 cm, real
• B. 4 cm, virtual
• C. 1.0 cm, real
• D. none of these

Answer 1

Answer: (A)

According to new cartesian sign convention, Object distance, u = -15 cm
Focal length of the concave mirror, f = -10 cm Height of the object, $h_O=2$ cm
According to mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}-\frac{1}{-15}=\frac{1}{-10}+\frac{1}{15}$
or $v=-30$ cm.
The image is formed at a distance of 30 cmfrom the mirror on the same side of the object.It is a real image.
Magnification of the mirror, $m=\frac{-v}{u}=\frac{h_1}{h_0}$
$\Rightarrow \frac{-\left(-30\right)}{-15}=\frac{h_I}{2}\Rightarrow h_1=-4$ cm.
Negative sign shows that image is inverted. The image is real, inverted, of size 4 cm at a distance of 30 cm in front of the mirror.