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You wish to project the image of a slide on a screen 9.00 \(\mathrm{m}\) from the lens of a slide projector. (a) If the slide is placed 15.0 \(\mathrm{cm}\) from the lens, what focal length lens is required? (b) If the dimensions of the picture on a \(35-\mathrm{mm}\) color slide are \(24 \mathrm{mm} \times 36 \mathrm{mm},\) what is the minimum size of the projector screen required to accommodate the image?

Short Answer

Expert verified
(a) The focal length needed is 14.7 cm. (b) The screen should be at least 144 cm x 216 cm.

Step by step solution

01

Understanding the Lens Formula

The lens formula relates the object distance \( u \), the image distance \( v \), and the focal length \( f \) of the lens: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]We aim to find the focal length \( f \) given \( u = 15.0 \text{ cm} \) (converted to meters: \( u = 0.15 \text{ m} \)) and \( v = 9.00 \text{ m} \).
02

Applying the Lens Formula

Substitute the known values into the lens formula:\[ \frac{1}{f} = \frac{1}{0.15} + \frac{1}{9.00} \]Calculate the sum of the fractions.
03

Solving for Focal Length

Performing the calculations:\[ \frac{1}{f} = 6.67 + 0.11 = 6.78 \]Thus,\[ f = \frac{1}{6.78} \approx 0.147 \text{ m} \text{ or } 14.7 \text{ cm} \].
04

Understanding Image Magnification

The magnification \( M \) of a lens is given by \( M = \frac{v}{u} \). Since we know \( v = 9.00 \text{ m} \) (or \( 900 \text{ cm} \)) and \( u = 15.0 \text{ cm} \), we can find the magnification as follows:\[ M = \frac{900}{15} = 60 \].
05

Calculating Screen Size

Given the dimensions of the slide \( 24 \text{ mm} \times 36 \text{ mm} \), we apply the magnification to find the image size on the screen. First, convert the dimensions to centimeters:\( 24 \text{ mm} = 2.4 \text{ cm} \) and \( 36 \text{ mm} = 3.6 \text{ cm} \).Using \( M = 60 \), multiply the original dimensions by \( 60 \):- Width: \( 2.4 \times 60 = 144 \text{ cm} \)- Height: \( 3.6 \times 60 = 216 \text{ cm} \).Thus, the minimum size of the projector screen required is \( 144 \text{ cm} \times 216 \text{ cm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Focal Length Calculation
Understanding how to find the focal length of a lens is essential when dealing with projection systems like a slide projector. The focal length is the distance between the center of a lens and its focus, where light converges to form a clear image. We use the lens formula to determine this distance:\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]where:
  • \( f \) is the focal length of the lens,
  • \( u \) is the distance from the object to the lens (object distance),
  • \( v \) is the distance from the image to the lens (image distance).
In the context of the slide projector problem, you're given these distances and you must solve for \( f \). Converting units as necessary, you substitute the object distance of 0.15 meters and the image distance of 9.00 meters into the formula. Solving, you find:\[f \approx 0.147 \text{ m} \text{ or } 14.7 \text{ cm}\]Understanding this helps ensure the lens you choose will focus light correctly to project a sharp image on the screen.
Image Magnification
Image magnification is a key concept in optics that describes how much larger (or smaller) an image is compared to the object's actual size. This is especially crucial for a slide projector, where you need to magnify the small slide into a screen-sized image.The formula for magnification \( M \) is:\[M = \frac{v}{u}\]where \( v \) and \( u \) are the image and object distances, respectively.In this scenario, the slide-to-lens distance is 15 cm, and the image-to-lens distance is 900 cm (or 9 meters). Plugging these values into the formula: \[M = \frac{900}{15} = 60\]This means the image on the projector screen will be 60 times larger than the slide.Understanding magnification helps in adjusting the lens correctly to achieve the desired image size.
Slide Projector
A slide projector is an optical device used to project images from a small slide onto a much larger screen. It functions by shining light through a small, transparent slide and using a lens to focus and enlarge the image onto a distant surface.
The core components include:
  • A bright light source to illuminate the slide,
  • A lens system to focus and magnify the image,
  • The slide holder that positions the slide between the light source and lens.
Efficient projection depends on correctly calculating the focal length and understanding magnification to ensure the image fits the screen size you require. A clear, correctly sized image on the screen highlights why knowing how to measure and set the physical distances between the slide, the lens, and the screen is crucial.
Optics
Optics is the scientific study of light behavior and its interaction with different media, crucial in understanding how a slide projector operates. It explores how light travels, reflects, refracts, and disperses through lenses. In a slide projector, optics principles apply when determining focal length and image quality. The lens in the projector uses refraction to converge light beams, thus focusing and enlarging the slide image on the screen. Key optics concepts include:
  • Refraction: Bending of light as it passes through the lens.
  • Focusing: Converging light rays to form a clear image at the projector's lens focal point.
  • Light Intensity: Affecting how well the image is illuminated.
A solid grasp of optics allows you to correctly adjust the components of a slide projector for optimal image display.

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Most popular questions from this chapter

Rear-View Mirror. A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 \(\mathrm{cm}\) (a) Another car is seen in this side mirror and is 13.0 \(\mathrm{m}\) behind the mirror. If this car is 1.5 \(\mathrm{m}\) tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?

The focal length of a simple magnifier is 8.00 \(\mathrm{cm} .\) Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 \(\mathrm{cm}\) in front of her eye? (b) If the object is 1.00 \(\mathrm{mm}\) high, what is the height of its image formed by the magnifier?

Three-Dimensional Image. The longitudinal magnification is defined as \(m^{\prime}=d s^{\prime} / d s .\) It relates the longitudinal dimension of a small object to the longitudinal dimension of its image. (a) Show that for a spherical mirror, \(m^{\prime}=-m^{2}\) . What is the significance of the fact that \(m^{\prime}\) is always negative? (b) A wire frame in the form of a small cube 1.00 \(\mathrm{mm}\) on a side is placed with its center on the axis of a concave mirror with radius of curvature 150.0 \(\mathrm{cm}\) . The sides of the cube are all either parallel or perpendicular to the axis. The cube face toward the mirror is 200.0 \(\mathrm{cm}\) to the left of the mirror vertex. Find (i) the location of the image of this face and of the opposite face of the cube; (ii) the lateral and longitudinal magnifications; (iii) the shape and dimensions of each of the six faces of the image.

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 \(\mathrm{cm}\) . An image of the 1.5 -tall coin is formed 6.00 \(\mathrm{cm}\) behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

You are in your car driving on a highway at 25 \(\mathrm{m} / \mathrm{s}\) when you glance in the passenger side mirror (a convex mirror with radius of curvature 150 \(\mathrm{cm}\) ) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.5 \(\mathrm{m} / \mathrm{s}\) when the truck is 2.0 \(\mathrm{m}\) away, what is the speed of the truck relative to the highway?

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