Problem 19 - College Physics

Question

A bottle \(6.0 \mathrm{~cm}\) tall is located \(75 \mathrm{~cm}\) from the concave surface of a mirror with a radius of curvature of \(50 \mathrm{~cm}\). Where is the image located, and what are its characteristics?

Step-by-Step Solution

Answer

The image is located 37.5 cm in front of the mirror, is real, inverted, and reduced.

Step 1: Understanding the Problem

We have a concave mirror with a bottle placed in front of it. The bottle's height is 6.0 cm, and it is located 75 cm in front of the mirror. The radius of curvature of the mirror is 50 cm. We need to find the location and characteristics of the image formed.

Step 2: Calculating the Focal Length

The focal length \(f\) of a concave mirror is half the radius of curvature \(R\). Given that \(R = 50 \mathrm{~cm}\), we can calculate \(f\) using the formula: \[ f = \frac{R}{2} = \frac{50}{2} = 25 \mathrm{~cm} \]

Step 3: Using the Mirror Equation

The mirror equation relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\). The mirror equation is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Given, \(d_o = 75 \mathrm{~cm}\) and \(f = 25 \mathrm{~cm}\), we substitute these values into the mirror equation to find \(d_i\).

Step 4: Solving for the Image Distance

Substitute the known values into the mirror equation:\[ \frac{1}{25} = \frac{1}{75} + \frac{1}{d_i} \] Rearranging the terms, we get:\[ \frac{1}{d_i} = \frac{1}{25} - \frac{1}{75} \]Calculating the right-hand side, we find:\[ \frac{1}{d_i} = \frac{3 - 1}{75} = \frac{2}{75} \]Taking the reciprocal gives:\[ d_i = \frac{75}{2} = 37.5 \mathrm{~cm} \] Thus, the image is located 37.5 cm in front of the mirror.

Step 5: Determining Image Characteristics

The characteristics of the image can be found using the magnification formula: \[ m = -\frac{d_i}{d_o} \]Substitute \(d_i = 37.5\) cm and \(d_o = 75\) cm:\[ m = -\frac{37.5}{75} = -0.5 \]The negative sign indicates the image is inverted, and since \(|m| < 1\), the image is reduced in size (smaller than the object). The image is real because it forms on the same side as the object.

Key Concepts

Mirror Equation

The mirror equation is a fundamental concept used to describe how light reflects off mirrors, specifically concave mirrors. This equation connects three important quantities:

• Object Distance (\(d_o\))
• Image Distance (\(d_i\))
• Focal Length (\(f\))

The formula itself is written as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]
This equation helps us calculate where an image will form when we know the object's distance from the mirror and its focal length. It's important to rearrange and manipulate the equation properly depending on what you need to find.
For instance, in our exercise, we were given the object distance and focal length, and we needed to find the image distance. By substituting the known values into the equation and rearranging, we can solve for the unknown.

Focal Length of Concave Mirror

The focal length of a mirror is a measure of how strongly it converges or diverges light. For a concave mirror, it is half of the radius of curvature (\(R\)). The formula is:\[f = \frac{R}{2}\]
In the given exercise, the radius of curvature was \(50\) cm, leading to a focal length of\(25\) cm. Understanding the focal length is crucial because it signifies the point where parallel light rays will converge after reflecting off the mirror's surface.
For concave mirrors, the focal point is located on the reflective side just inside the curve. This property is key in many applications, such as focusing beams in telescopes and in make-up mirrors to magnify one's image. By knowing the focal length, you can go ahead and use it in the mirror equation for further calculations.

Magnification Formula

The magnification formula provides insights into the size and orientation of an image formed by a concave mirror. The formula for magnification (\(m\)) is:\[m = -\frac{d_i}{d_o}\]

• A positive magnification implies an upright image.
• A negative magnification indicates the image is inverted.
• When \(|m| < 1\), the image is smaller than the object.
• When \(|m| > 1\), the image is larger.

In the described problem, the magnification came out to be \(-0.5\). This negative result tells us the image is inverted compared to the object. Also, since \(|m| < 1\), it confirms the image is reduced in size. Understanding magnification is pivotal to recognizing not only an image's size but also its orientation. These insights are vital in fields involving optics, engineering, and even art, allowing precise control over image projection and manipulation.