Thin lenses

The parallel beam of light will converge to a single point, known as the principal focus, on the other side of the lens. This point is located at a distance of 5.0 cm from the lens along the principal axis.

The parallel beam of light will diverge. The rays will appear to originate from a point on the same side of the lens as the incoming parallel beam. This point is the principal focus of the diverging lens.

The parallel light rays are refracted by the lens and converge to a single point. (1 mark)
Principal focus (focal point): The point on the principal axis where parallel rays of light converge after passing through a converging lens. (1 mark)

1. Draw the lens and principal axis.
2. Draw the object 6.0 cm from the lens.
3. Draw a ray from the top of the object parallel to the principal axis, refracting through the focal point on the other side of the lens.
4. Draw a ray from the top of the object passing through the center of the lens (this ray should be undeviated).
5. The point where the two rays intersect is the location of the top of the image. Draw the image.
Explanation: The intersection of the rays shows where the image is formed. This diagram should result in a real, inverted, and magnified image.

1. The image is inverted, while the object is upright.
2. The image can be magnified or diminished relative to the object's size.
Explanation: When an object is placed beyond the focal length of a converging lens, the image formed is real and inverted. Depending on the object distance relative to the focal length, the image can be larger or smaller than the object.

Enlarged: The image height (6.0 cm) is greater than the object height (2.0 cm).
Inverted: The question states that the image is inverted.
Real: An inverted image formed by a single lens is always a real image. This is because real images are formed by the actual intersection of light rays, which is necessary to create an inverted image.

1. Enlarged/Same size/Diminished: Refers to the size of the image compared to the object.
2. Upright/Inverted: Describes the orientation of the image relative to the object.
3. Real/Virtual: Indicates whether the image is formed by actual intersection of light rays (real) or the apparent intersection of light rays (virtual).

Virtual images are formed where diverging rays are extrapolated backwards. These rays do not actually converge at a single point in space, so they cannot form a visible projection on a screen. Only real images, formed by converging rays, can be projected.

1. A real image is formed by converging rays, while a virtual image is formed by diverging rays extrapolated backwards.
2. A real image can be projected onto a screen, while a virtual image cannot.

Ray Diagram:
1. Draw the lens and principal axis.
2. Draw the object 3 cm from the lens.
3. Draw a ray from the top of the object parallel to the principal axis. After refraction, it passes through the focal point on the other side of the lens (5 cm from the lens).
4. Draw a ray from the top of the object through the center of the lens. This ray continues in a straight line.
5. Trace both rays BACKWARDS. Where they intersect represents the top of the virtual image.

Image Location: The rays diverge after passing through the lens, so the image is formed on the SAME side of the lens as the object.
Image Nature: Virtual (because rays do not actually converge to form it).

1. Real images are formed by the actual intersection of light rays, whereas virtual images are formed by the apparent intersection of light rays (extensions of the rays).
2. Real images can be projected onto a screen, whereas virtual images cannot.
3. Real images are always inverted relative to the object (for a converging lens), while virtual images formed by a single converging lens are always upright.

Answer:

1. Virtual: Since the object is placed inside the focal length.
2. Upright: Virtual images formed by converging lenses are always upright.
3. Magnified: Since object is inside focal length.

To quantify the magnification: 1/f = 1/v + 1/u, 1/8 = 1/v + 1/-4, 1/v = 1/8 + 1/4 = 3/8, v = 8/3, which means that the magnification is M = v/u = (8/3)/4 = 2/3 (NOTE: error in previous statement that image is magnified, it's actually diminished since 2/3 < 1. The principle holds though that object is placed within the focal length)

Final Answer: Virtual, Upright, Diminished.

Answer:

1. The object must be placed inside the focal length (u < f).
2. The lens must be a converging lens (also known as a convex lens).
3. The eye must be positioned to view the virtual, magnified image formed by the lens.

Focal length, *f* = (image distance * object distance) / (image distance - object distance)

Image distance = -1.5 m (negative because the image formed by the lens needs to be at the person's near point)
Object distance = 0.25 m

*f* = (-1.5 m * 0.25 m) / (-1.5 m - 0.25 m)
*f* = -0.375 m² / -1.75 m
*f* = 0.21 m

A converging lens of focal length 0.21 m is needed. The image must be formed at the person's near point, which is why the image distance is negative.

In a short-sighted person, the eye lens focuses light from distant objects *in front* of the retina, resulting in a blurry image. A diverging lens is placed in front of the eye to spread out the light rays *before* they enter the eye. This shifts the focal point of the eye lens further back, so it now falls *on* the retina, resulting in a clear image. The diverging lens creates a virtual, upright and diminished image *closer* to the eye than the actual object.