An object is placed 30mm in front of a lens. An image of the object is located 90mm behind the lens.

a) Is the lens converging or diverging? explain.

b) What is the focal length of the lens?

c) Draw a diagram with lens at x=0, locate the image.

d) Is the image real or virtual?
Is it smaller, larger, or same size as the object?
Is it inverted or upright in comparison to the object?

e)the lens is replaced by a concave mirror with focal lenght 20 mm. Draw the mirror at position x=0 so that a real image is formed. Draw at least two rays and locate the image to show this situation.

f) If the object height is 15 mm, calculate the height of the image and the magnification of the mirror.

a) The lens is converging, because the object is placed in front of the lens and the image is located behind the lens.

b) The focal length of the lens is 60 mm.

c) The diagram would look like this:
Object: x = -30 mm
Lens: x = 0 mm
Image: x = 90 mm

d) The image is real, smaller than the object, and inverted in comparison to the object.

e) The diagram would look like this:
Object: x = -30 mm
Mirror: x = 0 mm
Image: x = -20 mm

f) The height of the image is 7.5 mm and the magnification of the mirror is 0.5.

a) The lens is converging. This is because the image formed is located behind the lens, indicating that the lens brings the rays of light together.

b) To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

where f is the focal length, v is the image distance, and u is the object distance. Plugging in the values given, u = -30mm (negative since the object is on the left side of the lens) and v = -90mm (negative since the image is on the right side of the lens), we get:

1/f = 1/-90 - 1/-30

Simplifying further,

1/f = -1/90 + 1/30

1/f = -2/90

1/f = -1/45

Therefore, the focal length of the lens is -45mm (negative indicating a converging lens).

c) To draw the diagram, place the lens at position x = 0. Mark the object at 30mm in front of the lens (to the left) and the image at 90mm behind the lens (to the right).

```
| |
| |
| |
| Lens |
------O---------O------
-30mm -0- 90mm
Object Image
```

d) The image formed by the lens is virtual since the rays of light do not actually intersect at the image location. The image is also smaller in size compared to the object. Lastly, the image is inverted relative to the object.

e) To draw the concave mirror at position x = 0, we would use a curved surface that bends inward. The rays of light would reflect off this surface. At least two rays can be drawn to locate the image formed:

- The first ray can be drawn parallel to the principal axis and reflect back through the focal point.
- The second ray can be drawn through the focal point and reflect back parallel to the principal axis.

```
Mirror
-xʘ-------->
| /
| /
| /
| /
| O
/|
/ |
/ |
/ |
/ |
| |
| |
| |
```
The rays will intersect at a point where the real image is formed behind the mirror.

f) To calculate the height of the image and the magnification of the mirror, we can use the mirror formula:

1/f = 1/v - 1/u

where f is the focal length, v is the image distance (negative since the image is formed on the same side as the object), and u is the object distance (positive since the object is on the opposite side of the lens). Plugging in the values, u = 30mm (positive since the object is to the left of the lens), and f = -20mm (negative since it is a concave mirror with the focus in front), we get:

1/-20 = 1/v - 1/30

Simplifying further:

1/-20 = 1/v - 1/30

-3/60 = 1/v - 1/30

-1/20 = 1/v - 1/30

Common denominator:

-3/60 = (3 - 2v) / 60v

-3 = 3 - 2v

-2v = -6

v = 3mm

The image is formed 3mm from the mirror. The height of the image can be calculated using the magnification formula:

magnification = image height / object height

Since the object height is 15mm, we need to determine the image height:

image height = magnification * object height

magnification = -v / u (negative due to the inverted image)

magnification = -3 / 30

magnification = -0.1

image height = -0.1 * 15

image height = -1.5mm

Therefore, the height of the image is -1.5mm (negative indicating an inverted image) and the magnification is -0.1.

a) To determine whether the lens is converging or diverging, we observe the behavior of light passing through the lens. If the lens causes the light rays to converge towards a focal point, it is a converging lens. On the other hand, if the lens causes the light rays to diverge, it is a diverging lens. In this case, since an image is formed behind the lens, the lens is a converging lens.

b) To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

where f is the focal length, v is the distance of the image from the lens, and u is the distance of the object from the lens.

Given that the object is located 30mm in front of the lens (u = -30mm) and the image is located 90mm behind the lens (v = -90mm), we can substitute these values into the formula:

1/f = 1/-90 - 1/-30

Simplifying:

1/f = -1/90 + 1/30
1/f = -2/90
1/f = -1/45

Taking the reciprocal of both sides:

f = -45mm

Therefore, the focal length of the lens is -45mm.

c) To draw a diagram with the lens at x = 0 and locate the image, we start by drawing a horizontal line representing the principal axis. Then, we draw a vertical line perpendicular to the principal axis at the position of the lens.

On the left side of the lens, we mark the object distance of 30mm (-30mm from the lens) from the lens. On the right side of the lens, we mark the image distance of 90mm (-90mm from the lens). We can represent the lens as a simple vertical line bisected by the principal axis.

d) The characteristics of the image can be determined by analyzing its properties:

- Real or virtual: Since the image is formed behind the lens, it is a real image.

- Size (larger, smaller, or the same): Since the image distance (-90mm) is greater than the object distance (-30mm), the image is larger than the object.

- Orientation (inverted or upright): Since the image is formed on the opposite side of the lens compared to the object, it is inverted in comparison to the object.

e) To draw the concave mirror at x = 0, we start by drawing a horizontal line representing the principal axis. Then, we draw a vertical line perpendicular to the principal axis at the position of the mirror.

Since the concave mirror forms a real image, we need to draw at least two rays:

- Incident ray parallel to the principal axis: We draw a ray coming from the object parallel to the principal axis, which reflects off the mirror and passes through the focal point.

- Incident ray passing through the focal point: We draw a ray coming from the object towards the focal point, which reflects off the mirror and becomes parallel to the principal axis.

The intersection point of these two rays will determine the location of the image formed by the concave mirror.

f) To calculate the height of the image and the magnification of the mirror, we can use the mirror formula:

1/f = 1/v - 1/u

where f is the focal length, v is the image distance, and u is the object distance.

Given that the focal length of the concave mirror is 20mm and the object distance is 30mm, we can substitute these values into the formula:

1/20 = 1/v - 1/30

Simplifying:

1/v = 1/20 + 1/30
1/v = 3/60 + 2/60
1/v = 5/60
1/v = 1/12

Taking the reciprocal of both sides:

v = 12mm

Now, to calculate the height of the image, we can use the magnification equation:

m = -v/u

where m is the magnification, v is the image distance, and u is the object distance.

Substituting the given values:

m = -12/30
m = -0.4

Therefore, the height of the image is -0.4 times the height of the object, indicating that the image is smaller in size.