Question

An object is $6\mathrm{cm}$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 virtual, located at $20\mathrm{cm}$behind of the lens, and it is inverted.

Step-by-step solution

Concepts and Principles

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

• Ray $1$is set diagonally to the major beam. This rays enters through to the main focus on the back focal plane since being light refracting by a lens.
• Ray $2$hits the lens as from centroid and goes in a linear fashion.
• Ray $3$is pulled out the front wall of the optic' center piece (or as if it were coming from the main focus if $s<f$) but arises transverse to the major axis.

  

Given Data

• The focal length of the lens is narrowing.
• The following is the object distance:$s=6\mathrm{cm}$
• The depth of field is as follows:$f=10\mathrm{cm}$

Required Data

We're required to do using light mapping to discover out if the picture is. So we should choose yet if the picture is true or simulated, straight or reversed.

Solution

The ray-tracing diagram shown in fig $1$. A number is inserted on the center of the lens. The focused points are selected and the optic is drawn. With all of its base also on central path, the thing is shown as an arrow. The three distinct rays detailed in the Premises and Foundations section were drawn out from tip of the arrow. The lights diverge from lens, or we may prolong the rays rearward to the point were they diverge using just a right angle. The image point is at this location.

We could see from the figure $1$that the object distance $s^{\mathrm{\prime }}=-15\mathrm{cm}$was indeed at which negative coefficient denotes also that picture is interactive. The picture also has seemed to be polarized.