a) See attachment
b) -180 cm, virtual image
c) Magnification: 10, image is 10 times the size of the object, upright
Explanation:
a)
The ray diagram for this situation is shown in attachment.
In order to find the position of the image, we proceed as follows:
- We draw a ray of light going from the tip of the object towards the lens, parallel to the principal axis - this ray is refracted towards the principal focus on the other side
- We draw another ray of light going from the tip of the object towards the centre of the lens
- We prolong both rays: we see that they don't meet on the right side of the lens. Therefore, we prolong them on the left side, where they meet - this means that the image is virtual, because it cannot be projected on a real screen (it is formed on the same side as the object).
b)
To find the nature and the position of the image, we use the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where:
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem:
[tex]f=20 cm[/tex] is the focal length
[tex]p=18 cm[/tex] is the object distance
Solving for q, we find the position of the image:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{20}-\frac{1}{18}=-0.00555 cm^{-1}\\q=\frac{1}{-0.00555}=-180 cm[/tex]
The negative sign indicates that the image is virtual (on the same side of the object).
c)
The magnification is given by:
[tex]M=-\frac{q}{p}[/tex]
where
q is the image distance from the lens
p is the object distance from the lens
Here we have
[tex]q=-180 cm[/tex]
[tex]p=18 cm[/tex]
So the magnification is:
[tex]M=-\frac{-180}{18}=10[/tex]
This means that the image size is a factor 10 times the object size. In fact, we can write
[tex]y'=My[/tex]
where
y' is the image size
y is the object size
Substituting,
[tex]y'=10y[/tex]
So, the image is 10 times the object (in size), and it has the same orientation, upright (because of the positive sign).
