Problem 34 - University Physics with Modern Physics

Question

A converging lens with a focal length of 7.00 \(\mathrm{cm}\) forms an image of a \(4.00-\mathrm{mm}\)-tall real object that is to the left of the lens. The image is 1.30 \(\mathrm{cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

Step-by-Step Solution

Answer

Object is at 5.35 cm and image at 17.39 cm; image is virtual.

Step 1: Understanding the Problem

We are dealing with a converging lens having a focal length of \( f = 7.00 \, \text{cm} \). An object of height \( h_o = 4.00 \, \text{mm} \) forms an image that is \( h_i = 1.30 \, \text{cm} \). We need to find the position of the object and image, and determine the nature (real or virtual) of the image.

Step 2: Use Magnification Formula

The magnification \( m \) is given by the formula \( m = \frac{h_i}{h_o} = \frac{-s_i}{s_o} \), where \( h_i \) and \( h_o \) are image and object heights, \( s_i \) is the image distance, and \( s_o \) is the object distance. Calculate \( m \):\[m = \frac{1.30 \text{ cm}}{0.40 \text{ cm}} = 3.25\]Since the image is erect, the magnification is positive, so \( m = 3.25 \).

Step 3: Relate Image and Object Distances

Since the magnification for an erect image is \( m = \frac{s_i}{s_o} \), we have \( \frac{s_i}{s_o} = 3.25 \), hence \( s_i = 3.25 s_o \).

Step 4: Use Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} \). Substitute \( s_i = 3.25 s_o \) into the lens formula to find \( s_o \):\[\frac{1}{7.00} = \frac{1}{s_o} + \frac{1}{3.25s_o}\]Combine the terms:\[\frac{1}{7.00} = \frac{4.25}{3.25s_o}\]Solving for \( s_o \) gives: \[s_o = \frac{3.25 \times 7.00}{4.25} = 5.35 \, \text{cm}\]

Step 5: Calculate Image Distance

Using \( s_i = 3.25 s_o \), calculate \( s_i \):\[s_i = 3.25 \times 5.35 \, \text{cm} = 17.39 \, \text{cm}\]

Step 6: Determine Nature of the Image

Since \( s_i \) is positive, the image is formed on the opposite side of the lens to the object, indicating a virtual image.

Key Concepts

Lens Formula

The lens formula plays a crucial role when dealing with lenses, especially when you want to determine the position of an image or object. It is expressed as:\[ \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} \]where:

• \( f \) is the focal length of the lens.
• \( s_o \) is the object distance from the lens.
• \( s_i \) is the image distance from the lens.

To find the image or object position, you only need two of these three parameters.
Using the formula involves substitution and sometimes rearranging the terms to solve for the unknown. In the case of a converging lens, the focal length and distances are often positive. Hence, the equation allows us to practically locate where the image will form based on the object’s position.
Remember, using the lens formula correctly can help you determine detailed characteristics of how light travels through a lens and forms an image.

Magnification

Magnification tells us how much larger or smaller an image is compared to the real object. It is defined by the formula:\[ m = \frac{h_i}{h_o} = \frac{-s_i}{s_o} \]where:

• \( h_i \) is the image height, and \( h_o \) is the object height.
• \( s_i \) is the image distance, and \( s_o \) is the object distance.

The magnification value can be either positive or negative.
If magnification is positive, this means the image is upright; if it is negative, the image is inverted.
In our example, the image was found to be upright and larger than the object, indicated by a positive magnification of 3.25. Understanding magnification helps determine the nature and orientation of the image formed by the lens.

Real and Virtual Images

Understanding the difference between real and virtual images is key when studying lenses. A real image is formed when light rays converge and actually meet at a point after passing through the lens. This kind of image can be projected on a screen as it is on the side opposite the object.
Virtual images, on the other hand, appear to be located on the same side as the object itself.

• With converging lenses, if the image distance \( s_i \) is positive, the image is real.
• If the image distance is negative, it indicates that the image is virtual.

In our case, since the object distance and image distance led to a positive \( s_i \) value, the image was noted as virtual, even though it appeared erect. This distinction is crucial for understanding where the image forms relative to the lens.