Finding Image Size for a Short Object in a Concave Mirror

When a short linear object is placed along the axis of a concave mirror, its image is also linear and lies along the axis. The size of this image is related to the size of the object by a factor called the axial magnification.

For a short object of length \(b\) placed along the axis, the size of the image \(b'\) is given by the absolute value of the change in image distance \(dv\) corresponding to the object length \(du\). If the object length is \(b\), then \(du = b\). The image length \(b'\) is approximately \(|dv|\).

We use the mirror formula relating the object distance \(u\), image distance \(v\), and focal length \(f\) of the concave mirror:

Mirror Formula

The mirror formula is given by:

\begin{equation*} \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \end{equation*}

Here, for a concave mirror, \(f\) is positive when using the convention \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) with distances \(u\) and \(v\) typically taken as positive for real objects and images (or using appropriate sign conventions).

Relating Object Distance, Image Distance, and Focal Length

We need to find a relationship between the distances and the focal length that can be used in the options provided. Let's rearrange the mirror formula to find an expression for the ratio of distances.

From the mirror formula, we can write \(\frac{1}{u}\) as:

\begin{equation*} \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \end{equation*}

Combining the terms on the right side:

\begin{equation*} \frac{1}{u} = \frac{v - f}{vf} \end{equation*}

Taking the reciprocal gives us an expression for \(u\):

\begin{equation*} u = \frac{vf}{v - f} \end{equation*}

Now, let's find the ratio \(\frac{u}{v}\):

\begin{equation*} \frac{u}{v} = \frac{\frac{vf}{v - f}}{v} \end{equation*}

\begin{equation*} \frac{u}{v} = \frac{vf}{v(v - f)} \end{equation*}

Cancelling \(v\) from the numerator and denominator (assuming \(v \ne 0\), which is true for image formation), we get:

\begin{equation*} \frac{u}{v} = \frac{f}{v - f} \end{equation*}

Squaring this ratio, we get:

\begin{equation*} \left( \frac{u}{v} \right)^2 = \left( \frac{f}{v - f} \right)^2 \end{equation*}

Image Size Derivation based on Options

For a short object lying along the principal axis, the image size \(b'\) is related to the object size \(b\) by the square of the ratio of image distance to object distance, i.e., \(b' = \left( \frac{v}{u} \right)^2 b\). However, the provided options include expressions involving the ratio \(\left( \frac{f}{v-f} \right)^2\).

We have found that \(\left( \frac{f}{v-f} \right)^2 = \left( \frac{u}{v} \right)^2\). One of the given options for the image size is \(\left( \frac{f}{v - f} \right)^2 b\).

Let's check Option 3:

Option 3: \(\left( \frac{f}{v - f} \right)^2 b\)

Based on our derivation, this expression is equivalent to \(\left( \frac{u}{v} \right)^2 b\).

Considering the form of the options provided, the size of the image can be expressed in terms of the focal length \(f\) and image distance \(v\) as given in Option 3.

Thus, the size of the image is \(\left( \frac{f}{v - f} \right)^2 b\).

This matches Option 3.

Concave Mirror Formulas Revision

Here are some key formulas for concave mirrors:

• Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
• Transverse Magnification (\(m_T\)): \(m_T = -\frac{v}{u}\)
• Axial Magnification (\(m_L\)) for a short object: \(m_L = \left| \frac{dv}{du} \right| = \left( \frac{v}{u} \right)^2\)

Using the mirror formula, the ratio \(\frac{v}{u}\) can also be expressed as:

• \(\frac{v}{u} = \frac{f}{u-f}\)
• \(\frac{v}{u} = \frac{v-f}{f}\)

And the ratio \(\frac{u}{v}\) can be expressed as:

• \(\frac{u}{v} = \frac{u-f}{f}\)
• \(\frac{u}{v} = \frac{f}{v-f}\)

The correct physical formula for image size of a short axial object is \(b' = \left(\frac{v}{u}\right)^2 b\). Substituting the relation \(\frac{v}{u} = \frac{f}{u-f}\) gives \(b' = \left(\frac{f}{u-f}\right)^2 b\), which matches Option 2. Substituting \(\frac{v}{u} = \frac{v-f}{f}\) gives \(b' = \left(\frac{v-f}{f}\right)^2 b\).

The provided Option 3, \(\left( \frac{f}{v-f} \right)^2 b\), corresponds to \(\left(\frac{u}{v}\right)^2 b\), because \(\frac{u}{v} = \frac{f}{v-f}\).

Additional Information on Axial Magnification in Optics

Axial magnification (\(m_L\)) describes how the length of a small object oriented along the optical axis is magnified by a mirror or lens. It is defined as the ratio of the length of the image (\(db'\) or \(dv\)) to the length of the object (\(db\) or \(du\)) in the limit as the object length approaches zero: \(m_L = \left| \frac{dv}{du} \right|\).

From the differentiation of the mirror formula, \(\frac{d}{du} \left( \frac{1}{v} + \frac{1}{u} \right) = \frac{d}{du} \left( \frac{1}{f} \right)\), we get \(-\frac{1}{v^2}\frac{dv}{du} - \frac{1}{u^2} = 0\), which leads to \(\frac{dv}{du} = -\frac{v^2}{u^2}\). Therefore, the axial magnification \(m_L = \left| -\frac{v^2}{u^2} \right| = \frac{v^2}{u^2}\). The image size \(b'\) of a short object of length \(b\) is \(b' = m_L \cdot b = \frac{v^2}{u^2} b\).

This confirms that the image size should be proportional to \(\left(\frac{v}{u}\right)^2\). The expression in Option 3, \(\left(\frac{f}{v-f}\right)^2 b\), is proportional to \(\left(\frac{u}{v}\right)^2 b\).