Problem 11 - College Physics

Question

A slide projector uses a lens of focal length 115 \(\mathrm{mm}\) to focus a 35 \(\mathrm{mm}\) slide (having dimensions 24 \(\mathrm{mm} \times 36 \mathrm{mm} )\) on a screen. The slide is placed 12.0 \(\mathrm{cm}\) in front of the lens. (a) Where should you place the screen to view the image of this slide? (b) What are the dimensions of the slide's image on the screen?

Step-by-Step Solution

Answer

Place the screen 276 cm from the lens; the image will be 552 mm by 828 mm.

Step 1: Identify the Given Data

We are given:- Focal length (\(f\)) of the lens: 115 mm = 11.5 cm- Object distance (\(d_o\)): 12.0 cm- Slide dimensions: 24 mm \(\times\) 36 mmWe need to find:(a) Image distance (\(d_i\)) from the lens to the screen.(b) Image dimensions on the screen.

Step 2: Use Lens Formula to Find Image Distance

The lens formula is given by:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]We can rearrange this to solve for the image distance \(d_i\):\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]Substitute the values:\[ \frac{1}{d_i} = \frac{1}{11.5} - \frac{1}{12.0} \]Calculating the above:\[ \frac{1}{d_i} = \frac{1}{11.5} - \frac{1}{12.0} = \frac{12 - 11.5}{138} = \frac{0.5}{138} \approx \frac{1}{276} \]Thus, the image distance \(d_i\) is approximately 276 cm.

Step 3: Calculate the Magnification

The magnification \(M\) is given by the formula:\[ M = \frac{d_i}{d_o} \]Substitute the values:\[ M = \frac{276}{12} = 23 \]The magnification is 23 times the size of the original slide.

Step 4: Determine the Image Dimensions

To find the dimensions of the image on the screen, use the magnification:- Width of the slide: 24 mm- Height of the slide: 36 mmCalculate the image dimensions:- Image width: \(24 \times 23 = 552\) mm- Image height: \(36 \times 23 = 828\) mmThus, the dimensions of the image on the screen are 552 mm by 828 mm.

Key Concepts

Focal Length

The focal length of a lens is the distance between the lens and the point where light rays converge to form a clear image. It is an essential characteristic of any lens, determining how much it can bend light. In this case, the projector uses a lens with a focal length of 115 mm, or 11.5 cm. This tells us how strongly the lens will converge parallel rays of light.

A shorter focal length means the lens is stronger and can focus light more sharply. Conversely, a longer focal length indicates a weaker lens with a less sharp focusing ability. When we calculate the image distance using the lens formula, we can see how this intrinsic property of the lens affects how far away the screen must be placed.

Image Distance

To find where to place the screen for a clear image, we need to calculate the image distance. The lens formula we use is:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)

In this equation:

• \(f\) is the focal length, which is 11.5 cm
• \(d_o\) is the object distance, in this problem, it's 12.0 cm
• \(d_i\) is the image distance, which we need to find

By rearranging the formula, we solve for \(d_i\):

• \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \)

Plugging in the numbers, we find \(d_i\) is approximately 276 cm.

This means the screen should be placed 276 cm from the lens to project a clear image of the slide.

Magnification

Magnification refers to how much larger or smaller an image is compared to the object being viewed. It's a crucial concept when determining the size of the projected image. Using the formula for magnification:

• \( M = \frac{d_i}{d_o} \)

This tells us how many times the image is magnified compared to the original object.

For this slide projector:

• \( d_i = 276 \) cm
• \( d_o = 12 \) cm
• Thus, \( M = \frac{276}{12} = 23 \)

The slide's image is magnified 23 times its actual size, making it much larger on the screen than in reality.

Image Dimensions

To calculate the dimensions of the image as it appears on the screen, we apply the magnification to the original dimensions of the slide. The slide's original dimensions in mm are:

• Width: 24 mm
• Height: 36 mm

Using the previously calculated magnification (23 times), we simply multiply the slide dimensions by this factor:

• Image width: \(24 \times 23 = 552\) mm
• Image height: \(36 \times 23 = 828\) mm

The projected image will, therefore, have dimensions of 552 mm by 828 mm on the screen.

This enlargement allows viewers to see details that would otherwise be too small to discern from the original slide.