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\(\bullet\) It's all done with mirrors. A photographer standing 0.750 \(\mathrm{m}\) in front of a plane mirror is taking a photograph of her image in the mirror, using a digital camera having a lens with a focal length of 19.5 \(\mathrm{mm}\) (a) How far is the lens from the light sensors of the camera? (b) If the camera is 8.0 \(\mathrm{cm}\) high, how high is its image on the sensors?

Short Answer

Expert verified
(a) 19.8 mm, (b) 1.056 mm (inverted)

Step by step solution

01

Mirror Image Distance

Since the photographer is 0.750 m in front of the plane mirror, her image appears the same distance behind the mirror. Thus, the image distance from the mirror is also 0.750 m.
02

Total Object Distance for Camera

To calculate the total distance between the camera and the image, we add the distance from the camera to the mirror and the distance from the mirror to the image. This gives us a total object distance of \(0.750 + 0.750 = 1.500\) m.
03

Convert Distances to Millimeters

Since the focal length is given in millimeters, we must convert the object distance to millimeters. Thus, \(1.500 \text{ m} = 1500 \text{ mm}\).
04

Use Lens Formula to Find Image Distance

To find the image distance of the camera's lens, we use the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, \]where \(f = 19.5 \text{ mm}\) is the focal length of the lens, \(d_o = 1500 \text{ mm}\) is the object distance and solve for \(d_i\) (the image distance):\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{19.5} - \frac{1}{1500}. \]By calculating this, we find that \(d_i \approx 19.8 \text{ mm}\).
05

Calculate Image Height Using Magnification

The magnification (\(M\)) of the lens is given by \[ M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}, \]where \(h_i\) is the image height on the sensors, and \(h_o = 80 \text{ mm}\) is the object's height. Substituting the known values:\[ M = -\frac{19.8}{1500} \approx -0.0132. \]Thus, \( h_i = M \cdot h_o = -0.0132 \times 80 \approx -1.056 \text{ mm}.\) The negative sign indicates the image is inverted.

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Key Concepts

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

Mirror Reflection
When you stand in front of a plane mirror, you see your image because of mirror reflection. A plane mirror reflects light back with the same angle at which it hits the surface. This means that the image appears as far behind the mirror as the object is in front of it.
For example, if you stand 0.750 meters away from a mirror, the image also appears 0.750 meters behind the mirror.
- A plane mirror creates a virtual image because the light rays don't actually come from the image. - This image is laterally inverted, meaning it is a mirror image of the object. - Unlike curved mirrors, plane mirrors do not magnify or reduce the size of the image.
Image Distance
Image distance refers to the distance between the lens of a camera or optical instrument and the image it forms. To find this distance, we often rely on key formulas, such as the lens formula. In the context of a mirror, the image distance is straightforward. The image distance is equal to the object distance from the mirror. When a camera is used to photograph this image, the total object distance to consider is the sum of the object-to-mirror distance and mirror-to-image distance.
For instance: - If you stand 0.750 meters from a mirror, your image also appears 0.750 meters from the mirror. - Therefore, for a camera placed at your position, the total distance to the image is 1.500 meters. - In a lens system, this must be converted to millimeters when working with focal lengths in millimeters.
Magnification
Magnification is the measure of how much larger or smaller an image is compared to the object. In optical systems like cameras, it's crucial for understanding how images scale.Magnification can be calculated with the formula:\[ M = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \]where:- \( M \) is the magnification,- \( h_i \) is the image height,- \( h_o \) is the object height,- \( d_i \) is the image distance,- \( d_o \) is the object distance.For a camera lens, if the calculated magnification is negative, this means the image will be inverted. In the case of the camera system in this example, with a magnification of approximately -0.0132, the image is smaller than the object and inverted.
Focal Length
Focal length is one of the most critical aspects of an optical lens. It defines how strongly the lens converges or diverges light.In a photography context:- The focal length tells us about the field of view and the magnifying power of the lens.- A shorter focal length provides a wider field of view, while a longer focal length gives a narrowed view but higher magnification.Lens formula:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:- \( f \) is the focal length,- \( d_o \) is the object distance,- \( d_i \) is the image distance.For example, in this exercise, the lens has a focal length of 19.5 mm. This is used to calculate the image distance when photographing the image seen in a mirror. Understanding focal length is vital for creating the desired visual effect in photographs.

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Most popular questions from this chapter

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?

A thin planoconvex lens has a radius of curvature of magnitude 22.5 \(\mathrm{cm}\) on the curved side. When a color chart is placed 48.0 \(\mathrm{cm}\) from the lens, green light of wavelength 550 \(\mathrm{nm}\) is focused 277 \(\mathrm{cm}\) from the lens and blue light of wavelength 450 \(\mathrm{nm}\) is focused 17 \(\mathrm{I} \mathrm{cm}\) from the lens. What are the indices of refraction for these two wavelengths of light?

\(\cdot\) A compound microscope has an objective lens of focal length 10.0 \(\mathrm{mm}\) with an eyepiece of focal length \(15.0 \mathrm{mm},\) and it produces its final image at infinity. The object to be viewed is placed 2.0 \(\mathrm{mm}\) beyond the focal point of the objective lens. (a) How far from the objective lens is the first image formed? (b) What is the overall magnification of this microscope?

\(\cdot\) A thin lens with a focal length of 6.00 \(\mathrm{cm}\) is used as a sim-ple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at infinity and that the lens is very close to the eye.

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