Problem 110 - Physics

Question

(a) For a diverging lens \((f=-20.0 \mathrm{~cm}),\) construct a ray diagram to scale and find the image distance for an object that is \(20.0 \mathrm{~cm}\) from the lens. (b) Determine the magnification of the lens from the diagram.

Step-by-Step Solution

Answer

The image distance is \(-10\ \mathrm{cm}\) and the magnification is \(0.5\).

Step 1: Understand the Problem

We have a diverging lens with focal length \(f = -20.0\ \mathrm{cm}\). The object is placed at a distance of \(d_o = 20.0\ \mathrm{cm}\) from the lens. We need to find the image distance \(d_i\) and magnification \(m\) using a ray diagram.

Step 2: Draw the Principal Axis and Lens Setup

Begin by drawing a straight horizontal line to represent the principal axis. Place the lens at the center of this axis and mark the focal points (F) at \(-20\ \mathrm{cm}\) on either side of the lens since it's a diverging lens. Now place the object at \(20\ \mathrm{cm}\) to the left of the lens on the principal axis.

Step 3: Draw Ray 1 - Parallel to the Principal Axis

Draw a ray from the top of the object parallel to the principal axis. In a diverging lens, the parallel ray appears to diverge from the focal point on the same side as the object. Extend the refracted ray backward through the lens to intersect the principal axis at the focal point.

Step 4: Draw Ray 2 - Through the Center of the Lens

Draw a second ray straight from the top of the object through the center of the lens. This ray will continue in a straight line, as the center of the lens does not bend light.

Step 5: Locate the Image

The two rays will not converge on the real side, but their backward extensions will meet on the same side as the object. Measure this intersection point from the lens along the principal axis to find the image distance \(d_i\). In this case, \(d_i = -10\ \mathrm{cm}\) (image is virtual and upright).

Step 6: Calculate Magnification

Magnification \(m\) is given by the equation \(m = -\frac{d_i}{d_o}\). Using the image distance \(d_i = -10\ \mathrm{cm}\) and object distance \(d_o = 20\ \mathrm{cm}\), calculate \(m = -\frac{-10}{20} = 0.5\). This indicates the image is half the size of the object and upright.

Key Concepts

Ray Diagram

To understand diverging lenses, ray diagrams are invaluable tools. They help us visualize how light rays interact with the lens to form an image. A diverging lens, or concave lens, spreads out light rays that are initially traveling parallel to its principal axis. The steps to create a ray diagram for a diverging lens include:

• Draw the principal axis, a straight horizontal line through the center of your work area.
• Place the lens in the center, marking the focal points on both sides. For a diverging lens, the focal points (F) are virtual, located on the same side as the object.
• Position the object, usually an arrow, on the principal axis on the left of the lens.
• Trace the first ray from the top of the object parallel to the principal axis. This ray diverges as if it originated from the focal point on the same side.
• Draw the second ray from the top of the object through the center of the lens. It continues in a straight line.

By extending the diverged ray backward, it appears to converge with the second ray, showing the virtual image location and behavior.

Image Distance

The image distance, represented as \(d_i\), is a measure of how far the image is from the lens. In lenses, these distances can be positive or negative:

• A positive image distance implies a real image, which is formed on the opposite side of the lens from the object.
• A negative image distance, like in our case, indicates a virtual image on the same side of the lens as the object.

To find this distance using a ray diagram:1. Locate the intersection of the backwards extensions of the diverging ray and the straight-through ray.2. Measure the distance from this point back to the lens along the principal axis.In our problem, the image distance is calculated to be \(-10\ \text{cm}\), affirming the presence of a virtual image.

Magnification

Magnification, noted as \(m\), tells us the size relationship and orientation between the image and object:

• It is calculated using the formula \(m = -\frac{d_i}{d_o}\), where \(d_i\) is the image distance and \(d_o\) is the object distance.
• A positive magnification indicates an upright image, while a negative value means it's inverted.
• The absolute value of magnification less than 1 means the image is smaller, while greater than 1 means larger.

Applying the formula, with \(d_o = 20\ \text{cm}\) and \(d_i = -10\ \text{cm}\), the magnification of this diverging lens is \(0.5\). This indicates the image is half the size of the object and upright.

Focal Length

The focal length of a lens is a crucial factor in determining how it converges or diverges light. For diverging lenses, the focal length \(f\) is negative, signaling that it spreads light rays apart.

• In the provided exercise, \(f = -20.0\ \text{cm}\) tells us the focal points are 20 cm from the lens.
• The negative sign reflects the nature of the diverging lens creating virtual images on the same side as the object.
• Knowing the focal length allows us to draw accurate ray diagrams, as it helps position the focal points which are integral for tracing ray paths.

Understanding the focal length is key to predicting how the lens will impact light paths and thus determine where and how an image will form.