Question

A short straight object of lengthLlies along the central axis of a spherical mirror, a distance pfrom the mirror. (a) Show that its image in the mirror has alength, ${\mathbf{L}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{L}\mathbf{(}\mathbf{f}\mathbf{/}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{f}\mathbf{)}{\mathbf{)}}^{\mathbf{2}}$(Hint: Locate the two ends of the object.) (b) Show that the longitudinal magnification is equal to${\mathbf{m}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{(}{\mathbf{L}}^{\mathbf{\text{'}}}\mathbf{/}\mathbf{L}\mathbf{)}$ is equal to m², where m is the lateral magnification.

Short answer

1. The image in the mirror has length $L\text{'}=L{\left(\frac{f}{\left(p-f\right)}\right)}^2$

2.The longitudinal magnification$m\text{'}=\frac{L\text{'}}{L}=m^2.$

Step-by-step solution

Listing the given quantities

Object distance p

Object length L

Hint: Locate the two ends of the object.

Understanding the concepts mirror equation

We use the mirror equation related to image distance, object distance and focal length. Using this relation, we can find the image distance. Using this, we calculated the image distance of the left and the right end of the object. Subtracting the initial distance and the final distance, we will get the image length in the mirror.

Formula:

$$\begin{gathered} \frac{1}{p}+\frac{1}{i}=\frac{1}{f} \\ m=-\frac{i}{p} \end{gathered}$$

Step 3: Calculations of the image in the mirror has length

(a)

Let us suppose the length of the object is L which lies along the central axis of the spherical mirror at a distance p from the mirror.

Let the object has two ends. If the right end of the object is at a distance p from the mirror, then the left end will be at distance p+L.

The image i₁ of the first right end is given by the mirror equation

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$

$i_1=\frac{fp}{p-f}$

For the left end which is located at p+L has the image distance,

$i_2=\frac{f(p+L)}{p+L-f}$

So, the length of the image formed in the mirror will be

$L\text{'}=i_1-i_2$

=$$\begin{gathered} \frac{fp}{p-f}-\frac{f(p+L)}{p+L-f} \\ =\frac{f{p}^2+fpL-f^{2}p-f{p}^2-fpL+f^{2}p+f^{2}L}{(p-f)(p+L-f)} \\ =\frac{f^{2}L}{(p-f)(p+L-f)} \end{gathered}$$

Neglecting the L of the second term of the denominator because the object is short by p-f.

role="math" localid="1663001144403" $$\begin{gathered} L\text{'}=\frac{f^{2}L}{(p-f)(p+L-f)} \\ =L{\left(\frac{f}{p-f}\right)}^2 \end{gathered}$$

The image in the mirror has length $L\text{'}=L{\left(\frac{f}{p-f}\right)}^2$

 Step 4: Calculations of the longitudinal magnification 

(b)

The lateral magnification is $m=-\frac{i}{p}$

But, $i=\frac{fb}{p-f}$

So, $m=-\frac{f}{p-f}$

Now, from the equation we proved in part (a)

$m\text{'}=\frac{L\text{'}}{L}$

$={\left(\frac{f}{p-f}\right)}^2$

= m²

The longitudinal magnification $m\text{'}=\frac{L\text{'}}{L}=m^2.$