Question

An object is $15\mathrm{cm}$in front of a diverging lens with a focal length of $-15\mathrm{cm}$. Use ray tracing to determine the location of the image. Is the image upright or inverted?

Short answer

The image is located at $7.5\mathrm{cm}$behind the lens and it is upright.

Step-by-step solution

Concepts and Principles

The three main rays are created from the tip of the thing to pinpoint the picture of a thin lens:

• Ray $1$is set diagonally to the central axis. This laser appears pointed far as from center piece in front side of the screen after becoming deflected by the focus.
• Ray $2$is brought forward towards the center piece on the outer focal plane and emerges parallel here to central axis as from aperture.
• Ray $3$enters the lens through the centre and continues in a straight line

Given Data

• The sight is skewed.
• The following seems to be the target range: $s=15\mathrm{cm}$
• The diffraction limit is as chooses to follow: $f=-15\mathrm{cm}$

Required Data

We're required to use laser mapping to establish out if the picture is. Thereafter, we can establish if the picture was erect or flattened.

Ray tracing 

The ray-tracing diagram is shown in Fig $1$. A scale is inserted on the optical axis. The focal parameters are defined and the lens is formed. With all of its base on the central path, the item is rendered as an arrow. The three specific rays specified in the Premises and Standards section are drawn first from tip of the arrow. The beams diverge from the lens, but we may continue the rays rearward to the point where the diverge using just a right angle. The picture point is at this place.

The real image is $s^{\mathrm{\prime }}=-7.5\mathrm{cm}$as is shown in the image, with the minus value reflecting that the picture is fictitious. The thing appears to be straight as well.