Question

An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 cm, calculate the image height

Answer 1

Answer:

2

Explanation:

In here if we are place object 5cm away from a convex lense that has focal length of 10cm. Image will be formed in same size of the object . And it's upright , virtual and magnified like normal hand lense. There for wee need to use lense equation to find the image distance to get magnification of the object

Magnification = image distance / object distance

M = V/U

By lense formula ,

1/v - 1/ u = 1/ f

1/v + 1/5 = 1/ 10 ( by adding sign convention )

V = -10 cm

Therefor magnification = V / U

= 10/5

= 2

Answer 2

Answer:

6 cm

Explanation:

The lens equation is :

$\frac{1}{f} =\frac{1}{d_{o}}+\frac{1}{d_{i}}$

where $f$ is the focal length, $d_{o}$ is the distance of the object and $d_{i}$ is the distance of the image.

To find the height of the image we first need to find the distance of the image $d_{i}$, so we clear for it in the last equation:

$\frac{1}{di} =\frac{1}{f} -\frac{1}{d_{o}}$

and since $f=10cm$ and $d_{o}=5cm$

$\frac{1}{di} =\frac{1}{10} -\frac{1}{5}$

$\frac{1}{di} =-\frac{1}{10}$

so

$d_{i}=-10$

Now to find the height of the image we use the magnification:

$M=\frac{h_{i}}{h_{o}} =-\frac{d_{i}}{d_{o}}$

where $h_{i}$ is the image height, and $h_{o}$ is the object height: $h_{o}=3cm$

clearing fot $h_{i}$:

$h_{i}=-\frac{h_{o}d_{i}}{d_{o}}$

substituting known values:

$h_{i}=-\frac{(3cm(-10cm))}{5cm}=\frac{30cm}{5}=6cm$

The image height is 6cm.