homework-and-exerciseslensesoptics
Q. An object of height 8 cm is placed in front of a lens. It's inverted image of height 4.8 cm is formed on the screen. If the focal length of the lens is 12 cm then by drawing at scale calculate the object distance, image distance and magnification.
Taking scale as 1cm = 4cm I've drawn the principal axis and the lens and marked focus (F) at 3cm from the lens. But how do I proceed from here with only the heights of the object and image?!
The lens equation and the ray tracing does agree. Let's denote the distance from the object and image distances to the first convex lens as $o_1$ and $i_1$ respectively. Let's denote the object and image distances to the second concave lens as $o_2$ and $i_2$. The focal lengths are $f_1=2 \text{ cm}$ and $f_2=-2 \text{ cm}$.
The lens equation reads $\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$.
For the first lens, $\frac{1}{3\text{ cm}}+\frac{1}{i_1}=\frac{1}{2\text{ cm}}$.
Solving for the image distance yields $i_1=6\text{ cm}$.
The image of the first lens becomes the object for the second lens, and since the two lenses are $4\text{ cm}$ apart, we have $o_2 = 4\text{ cm}-i_1 = -2\text{ cm}$.
The lens equation for the second lens reads as $\frac{1}{-2\text{ cm}}+\frac{1}{i_2}=\frac{1}{-2\text{ cm}}$.
Thus $1/i_2 = 0$, and $i_2=\infty$. There is neither a real image (converging rays in front of the second lens) nor a virtual image that is formed (diverging rays). Rather the rays are parallel.
Well, we can see this agrees with the ray tracing diagram (made with OpticalRayTracer software ver. 9.6):
The image could be real or virtual. We'll start with a real image. Also, we'll consider a point object and an ideal lens.
For a real image of a point to be formed, the rays emitted by or reflected from that point have to converge at some other point in space.
If a point (blue dot on the diagrams below) is placed in a focal plane of a convex lens and its rays, collected by the lens, are coming out parallel to each other, they, obviously, are not going to to converge and, therefore, are not going to form an image.
If a point is placed in front of the focal plane, the rays are going to converge and form a real image.
If a point is placed behind the focal plane (i.e. between the focal plane and the lens), the rays are going to diverge and, therefore are not going to form a real image. If the diverging rays are extended backwards, they will meet at some point (of the apparent divergence) behind the lens, forming a virtual image.
Hopefully, this clarifies the picture.
Best Answer
Do you know how to draw the image of an object set at a certain distance from the lens? It is sort of the inverse. Start by drawing the ray parallel to the optical axis from the object incident onto the lens and refracted through the focus $F$.
Extend both rays as much as you can (line in blue). Then, from the optical axis on the image side, find the perpendicular length of image that corresponds to the image length (the base is the focal axis and the tip of the image is where the image meets the refracted ray initially drawn).
Once you have it, join this point to the optical centre (the centre of the lens) and extend this ray (in purple) to meet the original incidence ray. Where they meet is where the object is.
You then have to measure the distances of the object, and of the image.