Question

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

Answer 1

Step 1, Given data

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

Step 2, Finding the position of the image

Using the lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Putting all the values

$\frac{1}{v}=\frac{1}{10}-\frac{1}{15}$

Or,

$v=30cm$

The 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.

Step 3, Finding the magnification

We know,

Magnification, $m=\frac{h^{\prime }}{h}=\frac{v}{u}$

Putting all the values

$\frac{h^{{}^{\prime }}}{2.0}=\frac{+30}{-15}=-2$

$h^{\prime }=-2\times 2=-4cm$

Magnification, $m=\frac{v}{u}=\frac{30}{-15}=-2$

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.