What is meant by the refraction of light? A converging lens is used as a magnifying glass - Leaving Cert Physics - Question 7 - 2006

To form a ray diagram:

1. Label the converging lens and focal points (F).
2. Place the object (closer than the focal length).
3. Draw rays from the object: one parallel to the axis which refracts through the focal point on the other side, and another passing through the center of the lens.
4. The rays diverge and appear to come from a virtual image that is upright and larger. Label this image clearly.

Diagram: (Insert a ray diagram showing these principles)

A diverging lens cannot create an upright, magnified image as it always produces a diminished (smaller) image. When light rays pass through a diverging lens, they spread out and do not converge, leading to a virtual image that is always smaller than the object.

Using the lens formula:

rac{1}{f} = rac{1}{u} + rac{1}{v}

Where:

• Focal length, $f = 8$ cm,
• Magnification m = rac{h'}{h} = rac{v}{u} = 4.

The two cases will be:

1. For real image, ( u = -10 ) cm and ( v = -40 ) cm;
2. For virtual image, ( u = -6 ) cm and ( v = -24 ) cm.

Using the formula for power of a lens: $P = P_1 + P_2$ Where:
$60 = 64 + P_1$, therefore $P_1 = 60 - 64 = -4$ m⁻¹.

Using the lens power formula: $P = \frac{1}{f}$ We found that $P = -4$ m⁻¹, hence: $f = \frac{1}{P} = \frac{1}{-4} = -0.25 \text{ m} = -25 \text{ cm}$

A diverging (concave) lens is used to correct this defect, which is known as short sight, short sightedness, or myopia.