Problem 36 - University Physics with Modern Physics

Question

A converging lens with a focal length of 12.0 \(\mathrm{cm}\) forms a virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) 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.

Step-by-Step Solution

Answer

The object is 68 cm left of the lens and 3.20 cm tall. The image is erect and on the same side as the object.

Step 1: Understand the Given Information

We are given a converging lens with a focal length \( f = 12.0 \, \text{cm} \), a virtual image height \( h'_i = 8.00 \, \text{mm} \) (convert this to cm for consistency, so \( h'_i = 0.800 \, \text{cm} \)), and image distance \( v' = 17.0 \, \text{cm} \). We need to find the object distance \( u \) and object height \( h_o \).

Step 2: Use the Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Since the image is virtual, the image distance \( v \) is negative. Substitute \( f = 12.0 \, \text{cm} \) and \( v = -17.0 \, \text{cm}. \)

Step 3: Calculate the Object Distance

Rearrange the lens formula to find \( u \): \( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \). Substitute \( v = -17.0 \, \text{cm} \) and \( f = 12.0 \, \text{cm} \), giving \( \frac{1}{u} = \frac{1}{-17} - \frac{1}{12} \). Calculate \( u \).

Step 4: Solve for Object Distance

Calculate \( \frac{1}{u} = \left(\frac{-12 - 17}{204}\right) \). Solve for \( u \) to get \( u \approx -68.0 \, \text{cm} \) (rounded to three significant figures).

Step 5: Use Magnification to Find Object Height

The magnification \( m \) is given by \( m = \frac{h'_i}{h_o} = -\frac{v}{u} \). Substitute \( v = -17.0 \, \text{cm} \) and \( u = -68.0 \, \text{cm} \) to find \( m \).

Step 6: Calculate the Object Height

Solve for \( h_o \) using \( m = \frac{h'_i}{h_o} \). Calculate \( m = -\frac{-17}{-68} = \frac{17}{68} = \frac{1}{4} \). Therefore, \( h_o = \frac{h'_i}{m} = \frac{0.800}{0.25} = 3.20 \, \text{cm} \).

Step 7: Analyze Image Characteristics

Since the magnification is positive and less than 1, the image is erect. The image being virtual also means that it appears on the same side of the lens as the object.

Key Concepts

Understanding the Lens Formula

The lens formula is a crucial tool for understanding how lenses form images. It's expressed as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where:

• \( f \) is the focal length of the lens,
• \( u \) is the object distance from the lens,
• \( v \) is the image distance from the lens.

For converging lenses, real images are typically formed on the opposite side of the object. However, when dealing with virtual images, as in this exercise, the image distance \( v \) is negative. This reversal helps determine the object's position and whether the image formed is virtual or real.

Deciphering Virtual Images

A virtual image is formed when the outgoing rays from a lens appear to diverge from a point. Virtual images are always upright and cannot be projected onto a screen. They occur in circumstances where the object is within the focal length of a converging lens.
In our example, the image's position being negative indicates a virtual image. Virtual images are always on the same side of the lens as the object, reinforcing their upright nature.

Exploring Magnification

Magnification tells us how much bigger or smaller an image is compared to the object. It's calculated as:

• \( m = \frac{h'_i}{h_o} = -\frac{v}{u} \)

Here, \( m \) is the magnification, \( h'_i \) is the image height, and \( h_o \) is the object height. When magnification is positive, like in our scenario, it implies the image is upright.
Since our calculated magnification is \( \frac{1}{4} \), the image is reduced in size compared to the object.

Creating a Ray Diagram

A ray diagram is a powerful visual tool that highlights how lenses create images. It uses principal rays to show the path light takes through a lens. In the case of a converging lens:

• A ray parallel to the principal axis refracts through the focal point on the other side.
• A ray passing through the center of the lens continues straight without refracting.
• A ray passing through the focal point becomes parallel to the principal axis after refraction.

These rays' divergence can be extended backward, illustrating where the virtual image forms. For this problem, the virtual image appears on the same side as the object, reinforcing the concept's understanding.