Answer to Question #200392 in Optics for Piolo

Question

An object of height 3.0 cm is placed 5.0 cm in front of a converging lens of focal length 20 cm and observed from the other side. Where and how large is the image?

di = ______ cm

hi = _______ m

Answer 1

Height of the object placed in front of lens is 3.0 cm.

The distance at which the object is placed in front of lens is 5.0 cm.

Focal length of the lens is 20.0 cm.

Write the expression for thin lens equation.

$\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o} \\ \frac{1}{d_i}=\frac{1}{f} - \frac{1}{d_o} \;(1)$

Here, f is the focal length of the thin lens, d_i is the distance of image from lens and d_o is the distance of object from lens.

Write the expression of magnification in terms of image distance to the ratio of object distance.

$m=\frac{-d_i}{d_o}\;(2)$

Magnification can also be defined as the ratio.

Write the expression of magnification in terms of height of image to the height of object.

$m=\frac{h_i}{h_o}\;(3)$

Compare equation (2) and (3) and rearrange the equation for magnitude of height of image.

$|h_i|= (\frac{d_i}{d_0})h_o\;(4)$

Here, m is the magnification of the lens, h_i is the height of image and h_o is the height of object.

The given lens is a converging lens. Therefore the focal length is taken positive.

Substitute 5.0 cm for d_o and 20 cm for f in equation (1).

$\frac{1}{d_i}=\frac{1}{20} -\frac{1}{5} \\ \frac{1}{d_i}=\frac{-3}{20} \\ d_i= -6.67 \;cm$

Substitute 6.67 cm for d_i, 5.0 cm for d_o and 3.0 cm for h_o in equation (4).

$|h_i|=( \frac{6.67 }{5.0 })3.0 cm=4.0 \;cm$

Thus, the image distance is 6.67 cm and formed in front of the lens and height of the image is 4.0cm.