Diverging Lens: Find Object Given Real Image & Size

[lha08]

Homework Statement

The focal length of a diverging lens is -20 cm. Locate the object, given that the image is real, erect, and 150% of the size of the object.

Homework Equations

(1/f)=(1/do)+(1/di)
M=-(di/do)

The Attempt at a Solution

i'm not really sure how a diverging lens forms a real image (i thought they could only from virtual since most of the images are on the same side) and a size that is larger than the original object. At first, I thought that the image would be inverted (negative magnification) since a virtual image would have a positive magnification (image is upright)...

 

[mgb_phys]

For a diverging lens revealed the image of the object is virtual, upright, reduced in size and located on the same side of the lens as the object.

You can form a real image from a diverging lens - but only by feeding it a converging beam from another lens (eg in a retrofocus wide angle camera lens)

 

[nlightner]

I can confirm that a diverging lens can indeed form a real image, even though it is not as common as a virtual image. This occurs when the object is placed between the focal point and the lens, and the image is formed on the same side as the object. In this case, the image is larger than the object, which is known as magnification.

To solve this problem, we can use the thin lens equation (1/f)=(1/do)+(1/di) to find the distance of the object from the lens (do). We can also use the magnification equation M=-(di/do) to find the distance of the image from the lens (di).

Substituting the given values, we have (1/-20)=(1/do)+(1/di) and M=-(di/do). Solving for do and di, we get do=-40 cm and di=-60 cm.

Therefore, the object is located 40 cm to the left of the lens and the image is located 60 cm to the left of the lens. Since the image is on the same side as the object, it is a real image. The fact that the image is 150% of the size of the object means that the magnification is M=-1.5, which is negative because the image is inverted.

In conclusion, a diverging lens can indeed form a real image, and in this case, the object is located 40 cm from the lens and the image is located 60 cm from the lens. The image is also inverted and 150% of the size of the object.