Question

The procedure known as ray tracing is a pictorial method for understanding image formation when lenses or mirrors are used. It consists of locating the image by the use of just three "special rays." The following tactics box explains this procedure for the case of a converging lens.

TACTICS BOX 23.2 Ray tracing for a converging lens

1. Draw an optical axis. Use graph paper or a ruler. Establish an appropriate scale.

2. Center the lens on the axis. Mark and label the focal points at distance f on either side.

3. Represent the object with an upright arrow at distance s. It is usually best to place the base of the arrow on the axis and to draw the arrow about half the radius of the lens.

4. Draw the three "special rays" from the tip of the arrow. Use a straightedge.

1. A ray parallel to the axis (Ray 1) refracts through the far focal point.

2. A ray that enters the lens along a line through the near focal point (Ray 2) emerges parallel to the axis.

3. A ray through the center of the lens (Ray 3) does not bend.

5. Extend the rays until they converge. This is the image point. Draw the rest of the image in the image plane. If the base of the object is on the axis, then the base of the image will also be on the axis.

6. Measure the image distance s'. Also, if needed, measure the image height relative to the object height.

Follow the steps above to solve the following problem: An object is 9.0 cm from a converging lens with a focal length of 2.8 cm. Use ray tracing to determine the location of the image.

Part A

The diagram below shows the situation described in the problem. The focal length of the lens is labeled f; the scale on the optical axis is in centimeters.

Draw the three special rays, Ray 1, Ray 2, and Ray 3, as described in the tactics box above, and label each ray accordingly. Do not draw the refracted rays.

Draw the rays from the tip of the object to the lens.

Answer 1

Concepts and reason

The concepts required to solve the given problem are method of ray tracing for converging lens, and lens equation.

First, describe the method of drawing a ray diagram and draw the ray diagram of the given object.

Then, check the result by finding the image distance using the lens equation.

Fundamentals

The points to consider while drawing the ray-diagram are as follows:

1.The ray parallel to the principle axis (Ray 1) passes through the focal point on the other side of the lens.

2.The ray passing through the focal point of the lens (Ray 2) emerges parallel to the principle axis.

3.The ray passing through the center of the lens (Ray 3) does not bend and pass undisturbed.

The expression of the lens equation is as follows:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Here, f is the focal length of lens, v is the image distance, and u is the object distance.

First choose the appropriate scale and mark the focal points both the sides. The focal point is 2.8 cm from the center of the lens and the object is 9.0 cm on the left side of the lens.

The ray 1 i.e. the incident ray from the object parallel to the principal axis passes through the focal point on the other side of the lens.

The ray 2 i.e. the ray passing through the focal point of the lens, emerges parallel to the principle axis.

The ray 3 i.e. the ray passing through the center of the lens, emerges without bending.

The following figure shows the ray diagram using the above described rays.

The expression of the lens equation is as follows:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Rearrange the above equation for v.

$\begin{array}{c}\\\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\\\\v = \frac{{fu}}{{u - f}}\\\end{array}$

Substitute 2.8 cm for u and 9.0 cm for u in the above equation.

$\begin{array}{c}\\v = \frac{{\left( {2.8{\rm{ cm}}} \right)\left( {9.0{\rm{ cm}}} \right)}}{{9.0{\rm{ cm}} - 2.8{\rm{ cm}}}}\\\\ = 4.06{\rm{ cm}}\\\\ \approx {\rm{4}}{\rm{.1 cm}}\\\end{array}$

Hence, the location of the image is 4.1 cm on the right side of the lens.

Ans:

The following figure shows the ray diagram.

The location of the image is 4.1 cm on the right side of the lens.