Problem 104 - Physics

Question

A slide projector has a converging lens whose focal length is \(105.00 \mathrm{~mm}\). (a) How far (in meters) from the lens must the screen be located if a slide is placed \(108.00 \mathrm{~mm}\) from the lens? (b) If the slide measures \(24.0 \mathrm{~mm} \times 36.0 \mathrm{~mm}\), what are the dimensions (in \(\mathrm{mm}\) ) of its image?

Step-by-Step Solution

Answer

a) 3.78 m, b) 23.3 mm x 35.0 mm.

Step 1: Understanding the Lens Equation

To find the distance where the image will form, we use the lens equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Here, \( f = 105.00 \mathrm{~mm} \) and \( d_o = 108.00 \mathrm{~mm} \).

Step 2: Calculating Image Distance

Rearrange the lens equation to solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \). Substitute the given values: \( \frac{1}{d_i} = \frac{1}{105} - \frac{1}{108} \). Solve to find \( d_i \).

Step 3: Converting Image Distance to Meters

Once \( d_i \) is obtained in millimeters, convert it to meters by dividing by 1000. This provides the screen distance in meters.

Step 4: Determining Image Magnification for Image Size

The magnification \( m \) is given by \( m = \frac{d_i}{d_o} \). Using the calculated \( d_i \) and given \( d_o = 108 \mathrm{~mm} \), find the magnification.

Step 5: Calculating Image Dimensions

Multiply the dimensions of the slide by the magnification to get the dimensions of the image. If the slide measures \( 24.0 \mathrm{~mm} \times 36.0 \mathrm{~mm} \), calculate the image dimensions as \( 24.0 \cdot m \) by \( 36.0 \cdot m \).

Key Concepts

Converging Lens

A converging lens is a specific type of lens that bends incoming parallel light rays to meet at a single point on the other side of the lens. This point is known as the focal point. Converging lenses are thicker at the center than at the edges, resembling a shape known as biconvex.
These lenses are commonly used in devices like cameras, glasses, microscopes, and projectors for their ability to focus light and create clear images.
Some important characteristics of converging lenses are:

• They can produce both real and virtual images depending on the object’s position.
• They can magnify objects, making them appear larger.
• The path that light takes through a lens is determined by its curvature and material.

Understanding how a converging lens works is essential, helping to predict how it affects the light passing through and consequently how the image appears.

Focal Length

The focal length of a lens is a critical concept that defines its optical power. It is the distance between the lens's center and its focal point, where light rays converge. For converging lenses, a shorter focal length means stronger lens power, allowing it to bend light more sharply.
The focal length is typically denoted by the symbol \( f \). In our exercise, we have a focal length of 105 mm. This value is essential for calculating where the image will form.

• A larger focal length gives a wider field of view, often producing a smaller, inverted image.
• Short focal lengths are used for close-up images, while longer ones are great for distant subjects.

The focal length can be found using the lensmaker's formula, which considers the radii of curvature of the lens surfaces and the refractive index of the lens material.

Image Magnification

Image magnification indicates how much larger or smaller an image is compared to the object itself. This is a dimensionless number given by the ratio of the image distance \( d_i \) to the object distance \( d_o \). Mathematically, magnification \( m \) can be expressed as: \[m = \frac{d_i}{d_o}\]In practical terms, if the magnification is greater than 1, the image is larger than the object. If less than 1, the image is smaller.
For our given lens setup with the object placed 108 mm away, calculating this ratio after determining \( d_i \) helps us understand the size difference compared to the original object.

• If \( m = 1 \), the image size equals the object size.
• If \( m > 1 \), the image is enlarged and if \( m < 1 \), the image is reduced.

Image magnification is vital in applications like photography and microscopy, where image clarity and size directly impact usability.

Image Distance Calculation

To determine the position of an image formed by a lens, we utilize the lens equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This relation connects the focal length \( f \), the object distance \( d_o \), and the image distance \( d_i \).
When calculating \( d_i \), rearrange the equation to: \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]By substituting the known values for \( f \) and \( d_o \), you can solve for \( d_i \). This result can then be converted from millimeters to meters by dividing by 1000, providing a practical way to measure where the projected image will land.

• Accurate calculation of \( d_i \) ensures the image appears sharply on the screen.
• The value of \( d_i \) can tell you if an image will appear on the screen or elsewhere.

This calculation process is crucial for designing optical systems accurately, such as projectors and cameras.