A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of object from the lens is 15 cm. Find the position, nature and size of the image. Calculate the magnification of the lens.
[4 Marks]

Solution:
Given
Object distance, $u=-15cm$
Focal length, $f=+10cm$
Object height, $h=+2cm$
Image distance, $v=?$
Image height, $h^{{}^{\prime }}=?$

Using the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}-\frac{1}{-15}=\frac{1}{+10}$
$v=30cm$ [1 Mark]

Positive sign of $v$ shows that the image is formed at a distance of 30 cm on the right side of the lens. Therefore, the image is real and inverted. [0.5 Marks]

From the formula of magnification,
$m=\frac{v}{u}=\frac{30}{-15}=-2$ [1 Mark]


Again, from the formula of magnification,
$m=\frac{h^{\prime }}{h}=\frac{v}{u}$
$\frac{h^{{}^{\prime }}}{2}=\frac{+30}{-15}=-2$ [0.5 Marks]
$h^{\prime }=-2\times 2=-4cm$ [0.5 Marks]

Negative sign with the magnification and height of the image shows that the image is inverted and real. Thus, a real image of height 4 cm is formed at a distance of 30 cm on the right side of the lens. Image is inverted and twice the size of the object. [0.5 Marks]