Problem 17 - Physics

Question

The image behind a convex mirror (radius of curvature \(=68 \mathrm{~cm}\) ) is located \(22 \mathrm{~cm}\) from the mirror. (a) Where is the object located and (b) what is the magnification of the mirror? Determine whether the image is (c) upright or inverted and (d) larger or smaller than the object.

Step-by-Step Solution

Answer

(a) Object is at 62.33 cm, (b) magnification is 0.353, (c) image is upright, (d) image is smaller than the object.

Step 1: Understand the Problem

We have a convex mirror with a radius of curvature of \( R = 68 \text{ cm} \). The image distance from the mirror is given as \( v = -22 \text{ cm} \) (negative because it's a virtual image behind the mirror). We need to find the object distance \( u \), the magnification \( m \), and determine the orientation and size comparison of the image.

Step 2: Calculate the Focal Length

The focal length \( f \) of a mirror is half its radius of curvature, but negative for a convex mirror. Hence, \( f = -\frac{R}{2} = -\frac{68}{2} = -34 \text{ cm} \).

Step 3: Use the Mirror Formula to Find Object Distance

The mirror formula is \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \). Plugging in \( f = -34 \text{ cm} \) and \( v = -22 \text{ cm} \), we get:\[ \frac{1}{-34} = \frac{1}{u} - \frac{1}{22} \].Solving for \( \frac{1}{u} \), we get:\[ \frac{1}{u} = \frac{1}{-34} + \frac{1}{22} = \frac{22 - 34}{748} = \frac{-12}{748} \],Thus, \( u = \frac{748}{12} = 62.33 \text{ cm} \).

Step 4: Calculate the Magnification

The magnification \( m \) is given by the formula \( m = -\frac{v}{u} \). Substitute \( v = -22 \text{ cm} \) and \( u = 62.33 \text{ cm} \):\[ m = -\left(\frac{-22}{62.33}\right) \approx 0.353 \].

Step 5: Determine the Image Orientation and Size

Since the magnification \( m \) is positive, the image is upright. Magnitude of \( m \) is \( 0.353 \), which is less than 1, indicating the image is smaller than the object.

Key Concepts

Radius of Curvature

The radius of curvature plays a vital role in understanding mirrors, especially convex mirrors. It represents the radius of the sphere from which the mirror segment is derived. In simpler terms, imagine the mirror as part of a large, hollow sphere. The radius of this sphere is the radius of curvature (\( R \)).
For this convex mirror example, the given radius of curvature is \( R = 68 ext{ cm} \). This value is crucial for determining the focal length of the mirror, which is half of this curvature.
It's important to note that for mirrors,

• Convex mirrors have a positive \( R \) for the geometrical sense, but we treat \( R \) as negative when calculating the focal length.
• The radius also influences the mirror formula, which helps in determining various distances related to the mirror.

Understanding this concept helps solve complex problems involving object and image distances.

Focal Length

The focal length of a mirror is the distance from the mirror to its focal point. It is a crucial factor in determining how light converges or diverges after reflecting from the mirror's surface. For any mirror, the focal length \( f \) relates directly to the radius of curvature \( R \).
Specifically, the formula is:

• For a spherical mirror: \( f = \frac{R}{2} \)
• For a convex mirror, the focal length is considered negative (\( f = -\frac{R}{2} \)).

In our example with \( R = 68 ext{ cm} \), the focal length is \( f = -34 ext{ cm} \). This negative sign indicates that a convex mirror diverges light rays. Understanding focal length helps in solving problems using the mirror formula, which ultimately relates object and image distances to the mirror.

Mirror Formula

The mirror formula connects the object distance (\( u \)), the image distance (\( v \)), and the focal length (\( f \)) of a mirror in a single equation:\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]This formula is fundamental in optics to predict where an image will form in relation to a mirror, allowing calculations of either the object or image distance if the other is known.
For a convex mirror in our scenario:

• Given \( f = -34 \text{ cm} \) and \( v = -22 \text{ cm} \)
• We solve the formula: \( \frac{1}{u} = \frac{1}{-34} + \frac{1}{22} \)

Finding \( u \) helps us understand how far an object needs to be placed from the mirror to form a particular type of image.

Image Magnification

Image magnification describes the ratio between the size of the image and the size of the object. In optics, it's a measure of how much larger or smaller the image is compared to the actual object. The magnification (\( m \)) of a mirror is given by the formula:\[ m = -\frac{v}{u} \]Here, \( v \) is the image distance, and \( u \) is the object distance.
In the problem, using \( v = -22 \text{ cm} \) and \( u = 62.33 \text{ cm} \):

• The magnification is \( m = 0.353 \)
• The positive value of \( m \) indicates the image is upright.
• The magnitude less than 1 (\( < 1 \)) suggests the image is smaller than the object.

Understanding image magnification helps us not only determine an image's size but also gives insight into its orientation, like determining if it's upright or inverted.