Question

An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.

Answer 1

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

$\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm$

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

$\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the heigth of the image

$h_o$ is the height of the object

$q=36.0 cm$

$p=16.0 cm$

Solving the formula for $h_i$, we find

$h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm$

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of $h_i$ is positive, the image is erect

- When the sign of $h_i$ is negative, the image is inverted

In this case, $h_i$ is negative, so the image is inverted.