Thin Lenses - 3.2.3 Virtual Images with Converging Lenses

This section explains how to draw and use ray diagrams to understand the formation of virtual images by converging lenses. A virtual image is an image that cannot be projected onto a screen; it appears to be behind the lens.

Key Concepts

• Converging Lens: A convex lens that bends light rays inwards.
• Object Distance (u): The distance of the object from the lens.
• Image Distance (v): The distance of the image from the lens.
• Focal Point (f): The point where parallel rays of light converge after passing through the lens.
• Object Position: For virtual images, the object is placed between the focal point and the lens.
• Image Characteristics: Virtual images formed by converging lenses are always upright and magnified.

Ray Diagram Construction

To draw a ray diagram for a virtual image formed by a converging lens:

1. Mark the Lens: Draw a curved line representing the converging lens. Mark the principal axis and the focal point (f) on this line.
2. Position the Object: Mark the position of the object (o) on the principal axis, such that it lies between the focal point (f) and the lens.
3. Draw the Rays: Draw three rays originating from the top of the object:
   • Ray 1: A ray parallel to the principal axis passes through the focal point (f).
   • Ray 2: A ray passing through the focal point (f) continues straight on.
   • Ray 3: A ray passing through the optical centre (centre of the lens) does not diverge.
4. Locate the Image: The image is formed where the extensions of the refracted rays meet. This intersection point is where the virtual image is located.
5. Measure Image Distance: Measure the distance between the lens and the image point. This is the image distance (v).

Ray Diagram Example

Lens Formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) of the lens:

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

For virtual images, the image distance (v) is negative.

Magnification

The magnification (m) of a lens is the ratio of the height of the image (h') to the height of the object (h):

$$ m = \frac{h'}{h} = -\frac{v}{u} $$

For virtual images, the magnification is negative, indicating that the image is inverted (although it appears upright). The absolute value of the magnification indicates the size of the image relative to the object.

Example Calculation

A converging lens has a focal length of 20 cm. An object is placed 10 cm to the left of the lens. Calculate the image distance and the magnification.

Given: $f = 20 \, cm$, $u = -10 \, cm$ (negative because the object is to the left of the lens)

Using the lens formula:

$$ \frac{1}{20} = \frac{1}{-10} + \frac{1}{v} $$ $$ \frac{1}{v} = \frac{1}{20} + \frac{1}{10} $$ $$ \frac{1}{v} = \frac{1}{20} + \frac{2}{20} = \frac{3}{20} $$ $$ v = \frac{20}{3} \, cm \approx 6.67 \, cm $$

The image distance is approximately 6.67 cm to the right of the lens. Since the image distance is positive, the image is virtual.

Calculate the magnification:

$$ m = -\frac{v}{u} = -\frac{20/3}{-10} = \frac{20}{30} = \frac{2}{3} $$

The magnification is 2/3. This means the image is 2/3 the size of the object and appears upright.

Summary

By understanding the ray diagrams and the lens formula, we can analyze the formation of virtual images by converging lenses. These images are always upright and magnified, and their position and size can be determined using these tools.