Question

A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

Short answer

Object is 7.03 cm left; image is erect, same side as object. Approx. 0.33 cm tall.

Step-by-step solution

Use the Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. Since it is forming a virtual image on the side of the light source, \( v = -17.0 \) cm, and \( f = 12.0 \) cm. Substituting these into the formula:\[\frac{1}{12} = \frac{-1}{17} - \frac{1}{u}\]Solve for \( u \) to find the position of the object.

Calculate Object Distance

Solving the equation \( \frac{1}{12} = \frac{-1}{17} - \frac{1}{u} \) gives:\[\frac{1}{u} = \frac{-1}{17} - \frac{1}{12}\]Finding a common denominator (204) for fractions:\[\frac{1}{u} = \frac{-12}{204} - \frac{17}{204} \Rightarrow \frac{1}{u} = \frac{-29}{204}\]Thus, \( u = \frac{204}{-29} \approx -7.03 \) cm. The object is located approximately 7.03 cm to the left of the lens.

Calculate Magnification and Object Size

The magnification \( m \) is given by \( m = \frac{v}{u} = \frac{\text{image height}}{\text{object height}} \). We know the image height is 8.00 mm (or 0.8 cm). \( m = \frac{-17}{-7.03} \approx 2.42 \). Using the magnification to find object height:\[m = \frac{0.8}{\text{object height}} \Rightarrow \text{object height} = \frac{0.8}{2.42} \approx 0.33 \text{ cm}\]The object is approximately 0.33 cm tall.

Determine Erect or Inverted Image and Same or Opposite Side

The magnification is positive, which means the image is erect. Since both the object and image distances are negative, they are on the same side of the lens as the object.

Draw the Principal-Ray Diagram

In a principal-ray diagram for a converging lens with a virtual image, draw a lens and an object on the left of the lens. Three principal rays are essential: - A ray parallel to the principal axis refracts through the lens and diverges as if it comes from the focal point on the left. - A ray through the center of the lens passes straight without deviation. - A ray appearing to come from the focal point on the right refracts parallel to the principal axis. These will produce a virtual, erect, and magnified image on the left of the lens, where the rays appear to converge.

Key concepts

Converging Lens

A converging lens is a fascinating optical element designed to bend incoming parallel light rays inward, causing them to converge at a focal point on the opposite side of the lens.
This type of lens is often thicker in the middle and thinner at the edges.
Converging lenses are typically used in applications that require magnification, such as microscopes and telescopes.
In optics, the focal length of the lens is a critical parameter that defines how strongly the lens can converge light.
For instance, a shorter focal length indicates a lens that bends the light more sharply. Converging lenses not only create real images but also virtual images, depending on the object's position in relation to the focal point.

• They can construct both real and virtual images.
• Real images are formed on the opposite side of the lens from the object.
• Virtual images form on the same side as the object.

Virtual Image

In the world of optics, a virtual image is intriguing because it appears to be located at a location from which light does not actually come.
This type of image cannot be projected onto a screen; it is visible only when looking through a lens or a mirror.
A converging lens creates a virtual image when the object is positioned closer to the lens than the focal length.
Such images are typically upright and magnified, making them useful in devices like magnifying glasses or simple microscopes.

• Virtual images appear erect (upright).
• They are formed on the same side of the lens as the object.
• Cannot be captured on a screen.

Lens Formula

The lens formula is a fundamental equation in optics that connects the object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) of a lens.
It is expressed as:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]
This formula becomes especially useful when determining unknown parameters if any two of the three variables are known.
By applying the lens formula, one can accurately predict the nature and size of the image formed by the lens given the position of the object.
Importantly, sign conventions play a crucial role:

• Distances measured against the direction of the incoming light are negative.
• Distances measured in the same direction as the incident light are positive.
• The formula helps deduce whether an image is virtual or real based on the signs of the distances.

Magnification

Magnification in optics determines how much larger or smaller the image is compared to the object.
It is given by the ratio of the height of the image (\( h' \)) to the height of the object (\( h \)), as well as the image distance to the object distance:\[m = \frac{h'}{h} = \frac{v}{u}\]
The magnification is not just about size; it also tells us about the orientation of the image.

• A positive magnification indicates an erect image.
• A negative magnification signifies an inverted image.
• Magnification can offer insights on whether the image is real (negative value of \( v \)) or virtual (positive value of \( v \)).

Understanding magnification allows you to predict how a lens will visually impact the appearance of objects.

Principal-Ray Diagram

Creating a principal-ray diagram is an invaluable skill in understanding how images form through lenses.
This diagram visually illustrates the path light rays take as they pass through a lens.
For a converging lens producing a virtual image, there are three principal rays to consider:

• A ray parallel to the principal axis will bend and diverge as though it originates from the focal point on the opposite side.
• A ray passing through the center of the lens remains undeviated.
• A ray seeming to come from the focal point on the lens's far side will recline parallel to the principal axis once it exits.

By drawing these rays correctly, you can construct a precise representation of where, and how, the virtual image is formed.
This process is crucial for visually confirming the behavior predicted by equations, enhancing one’s comprehension of lens optics.