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An object 0.600 \(\mathrm{cm}\) tall is placed 16.5 \(\mathrm{cm}\) to the left of the vertex of a concave spherical mirror having a radius of curvature of \(22.0 \mathrm{cm} .\) (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

Short Answer

Expert verified
The image is 27.5 cm from the mirror, inverted, real, and 1.002 cm tall.

Step by step solution

01

Understanding Mirror Properties

First, note that the object is placed in front of a concave mirror. The radius of curvature of the mirror is given, which is 22.0 cm. This means the focal length \( f \) can be calculated using the formula \( f = \frac{R}{2} \), where \( R \) is the radius of curvature. Thus, \( f = \frac{22.0\, \text{cm}}{2} = 11.0\, \text{cm} \).
02

Ray Diagram for Mirror

Draw a concave mirror and represent the principal axis. Place the object 16.5 cm to the left of the mirror. Sketch the principal rays: 1) A ray parallel to the principal axis that will reflect through the focal point, 2) A ray through the focal point that reflects parallel to the principal axis, and 3) A third ray through the center of curvature reflecting back on itself. The intersection of reflected rays gives the image location.
03

Apply Mirror Formula

Use the mirror formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o = 16.5\, \text{cm} \) is the object distance and \( f = 11.0\, \text{cm} \). Substitute the known values to find the image distance \( d_i \): \[ \frac{1}{11.0} = \frac{1}{16.5} + \frac{1}{d_i} \].
04

Solve for Image Distance

Re-arrange and solve the equation: \[ \frac{1}{d_i} = \frac{1}{11.0} - \frac{1}{16.5} \]. Calculate to find \( d_i \approx 27.5\, \text{cm} \). The positive value indicates that the image forms on the same side as the object, making it a real image.
05

Determine Image Size and Orientation

Calculate the magnification \( m \) using \( m = -\frac{d_i}{d_o} \). Substitute \( d_i = 27.5\, \text{cm} \) and \( d_o = 16.5\, \text{cm} \): \[ m = -\frac{27.5}{16.5} \approx -1.67 \]. The negative magnification suggests the image is inverted. To find the image size, \( h_i = m \times h_o = -1.67 \times 0.6\, \text{cm} \approx -1.002\, \text{cm} \).
06

Summary of Image Characteristics

The image is located approximately 27.5 cm to the right of the mirror. It is 1.002 cm tall, inverted, and real, as it forms on the same side as the object with a negative magnification.

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Key Concepts

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

Spherical Mirror
A spherical mirror is a mirror whose reflecting surface is part of a sphere. Concave spherical mirrors have their reflective surface on the inner side, resembling a bowl, while convex mirrors have it on the outer side. In our exercise, we are dealing with a concave spherical mirror. These mirrors have specific properties, such as a center of curvature and a radius of curvature, both of which define the geometry of the mirror. The radius of curvature is the radius of the sphere from which the mirror segment is taken. It plays a critical role in determining the focal length, which is half the radius of curvature. Therefore, for a mirror with a radius of curvature of 22.0 cm, the focal length is calculated as \( f = \frac{22.0\, \text{cm}}{2} = 11.0\, \text{cm} \). Knowing the focal length helps us predict how the mirror will form images of objects placed in front of it.
Mirror Equation
The mirror equation provides a mathematical relationship between the object distance \( d_o \), the image distance \( d_i \), and the focal length \( f \) of a spherical mirror. The formula is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). This equation is central to understanding how different factors affect the formation and characteristics of images. By substituting known values into the equation, one can calculate the missing variable.
In the case of our exercise, we have \( d_o = 16.5\, \text{cm} \) and \( f = 11.0\, \text{cm} \). Plugging these into the mirror equation allows us to solve for \( d_i \), which is the image distance. Rearranging the formula to \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \), and calculating gives us an image distance of approximately 27.5 cm. A positive \( d_i \) implies that the image is real and located on the same side as the object.
Ray Diagram
A ray diagram is a practical tool for visually understanding how an image forms with spherical mirrors. It involves sketching principal rays to locate where they intersect, which determines the image position. For a concave mirror, common rays to draw include:
  • A ray parallel to the principal axis, which reflects through the focal point.
  • A ray passing through the focal point, which reflects parallel to the principal axis.
  • A ray that goes through the center of curvature, reflecting back on itself.
These rays come together at a point to form the image of the object. In our example, the intersection indicates that the image forms on the same side of the mirror as the object, consistent with our findings using the mirror equation. This visual approach complements mathematical calculations, offering an intuitive grasp of image formation.
Image Characteristics
Image characteristics, such as position, size, orientation, and type, are key outcomes when studying image formation by spherical mirrors. Using the mirror equation and magnification formula, we can describe these features in detail. The image position is found to be 27.5 cm from the mirror’s vertex, indicating it's on the same side as the object, making it a real image.
The magnification \( m \) is calculated using \( m = -\frac{d_i}{d_o} \), which for this exercise results in \( m \approx -1.67 \). A negative magnification value implies that the image is inverted relative to the object. The size of the image \( h_i \) can be found by multiplying the magnification \( m \) by the object height \( h_o \), yielding \( h_i \approx -1.002\, \text{cm} \). Hence, the image is larger than the object, real, inverted, and located on the same side of the mirror.

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Most popular questions from this chapter

Recall that the intensity of light reaching film in a camera is proportional to the effective area of the lens. Camera \(A\) has a lens with an aperture diameter of 8.00 \(\mathrm{mm}\) . It photographs an object using the correct exposure time of \(\frac{1}{30} \mathrm{s}\) . What exposure time should be used with camera \(B\) in photographing the same object with the same film if this camera has a lens with an aperture diameter of 23.1 \(\mathrm{mm} ?\)

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