Question

An object is $20\mathrm{cm}$in front of a converging lens with a focal length of $10\mathrm{cm}$. Use ray tracing to determine the location of the image. Is the image upright or inverted?

Short answer

The image is located at$20\mathrm{cm}$ in front of the lens and it is inverted.

Step-by-step solution

Concepts and Principle

The three main rays were drawn as from top of thing to pinpoint the picture of a diffracted beam:

• Ray $1$is depicted orthogonal to the major beam. This laser enters through into the center piece on the opposite side of the screen after just being birefringent by the lenses.
• Ray $2$leaves the lens out from centroid as well as continues in either a straight line.
• Ray $3$is attracted and through main focus on the front focal plane or as if arriving out from main focus if and rises up as from lens similar to the center point.

Given Data

• The focal length of the lens is narrowing.
• $s=20\mathrm{cm}$is the object distance.
• $f=10\mathrm{cm}$is the focal length.

Required Data

We're intended to use light tracing to establish out if the picture is. After that, we must assess whether the visual is straight or flipped.

Three special ray 

The ray-tracing diagram is shown in Fig $1$. A measure is inserted on the optical axis. The focal points or the lens is formed. With its basis on the optical axis, the item is portrayed as an arrow. The three special rays discussed in the Concepts and Philosophies portion are inspired as from tip of the arrow. The image's tip is drawn out from conjunction of the $3$rays.

The picture distance is $s^{\mathrm{\prime }}=20\mathrm{cm}$, as shown in the graphic in Figure $1$. In addition, the image looks to be inverted.