A luminous object and a screen are a fixed distance D apart. (a) Show that a converging lens of focal length f, placed between object and screen, will form a real image on the screen for two lens positions that are separated by a distance$\mathbf{d}\mathbf{=}\sqrt{\mathbf{D}\mathbf{(}\mathbf{D}\mathbf{-}\mathbf{4}\mathbf{f}\mathbf{)}}$(b) Show that ${\left(\frac{D-d}{D+d}\right)}^{\mathbf{2}}$gives the ratio of the two image sizes for these two positions of the lens.

1. Focal length is f.
2. Object distance is p.
3. Image distance is i.

Using the lens equation, solve for object distances x₁ and x₂. The magnification factor at the given distances x₁ and x₂is calculated and the magnification ratio is obtained bt finding the individual magnification.

Consider the lens formula as follows

$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$ ….. (i)

The magnification formula of the lens as follows:

$\left|m\right|=\frac{i}{p}$ …… (ii)

Consider the formula for roots of the quadratic equation as follows:

role="math" localid="1663081096118" $x=\frac{(-b\pm \surd (b^2-4ac\left)\right)}{2a}$ …… (iii)

A luminous object and screen are a fixed distances D apart. Consider the object distance to be x; then the distance between the object and image is D-x. Thus, from the lens equation (i), we can get that

$\frac{1}{x}+\frac{1}{(D-x)}=\frac{1}{f}$

By multiplying fx (D-x) for both the sides in the above equation and the solve,

$$\begin{gathered} f(D-x)+fx=x(D-x) \\ fD-fx+fx=xD-x^2 \\ fD=xD-x^2 \\ {x}^2-xD+fD=0 \end{gathered}$$

The above is a quadratic equation, thus, solving for x, we get two values x₁ and x₂. The both roots can be given using equation (iii) as follows:

$$\begin{gathered} x_1=\frac{D-\surd D^2-4Df\left)\right)}{2} \\ =\frac{D-\sqrt{D(D-4f)}}{2} \end{gathered}$$

Also:

$$\begin{gathered} x_2=\frac{D+\surd D^2-4Df\left)\right)}{2} \\ =\frac{D+\sqrt{D(D-4f)}}{2} \end{gathered}$$

The distance between the object positions now using the above values can be given as:

d = x₂- x₁

Substitute the values and solve as:

$$\begin{gathered} d=\frac{D+\sqrt{D(D-4f)}}{2}-\frac{D-\sqrt{D(D-4f)}}{2} \\ =\frac{D+\sqrt{D(D-4f)}-D+\sqrt{D(D-4f)}}{2} \\ =\frac{2\sqrt{D(D-4f)}}{2} \\ =\sqrt{D(D-4f)} \end{gathered}$$

Hence, the separation distance is $\sqrt{D(D-4f)}$.

The ratio of the image sizes is the same as the ratio of the lateral magnifications. If the object is at p=x1, the magnitude of the lateral magnification is given using equation (ii) as follows:

$\left|m_1\right|=\frac{i_1}{p_1}$=$\frac{D-x_1}{x_1}$

Now, the x₁ value is given as:

$$\begin{gathered} x_1=\frac{D-\sqrt{D(D-4f)}}{2} \\ =\frac{D-d}{2} \end{gathered}$$

Thus, the lateral magnification is given using the above value in for |m1| as:

$$\begin{gathered} \left|m_1\right|=\frac{D-x_1}{x_1} \\ =\frac{{\displaystyle \frac{D+d}{2}}}{{\displaystyle \frac{D-d}{2}}} \\ =\frac{{\displaystyle D-\frac{D}{2}+\frac{d}{2}}}{{\displaystyle \frac{D-d}{2}}} \end{gathered}$$

$=\frac{D-{\displaystyle \frac{D-d}{2}}}{{\displaystyle \frac{D-d}{2}}}$

Solve further as:

$\left|m_1\right|=\frac{D+d}{D-d}$

Similarly, when object is at x₂, the lateral magnification can be given using equation (ii) as follows:

$\left|m_2\right|=\frac{i_2}{p_2}$

Now, the x₂ value is given as:

$$\begin{gathered} x_2=\frac{D+\sqrt{D(D-4f)}}{2} \\ =\frac{D+d}{2} \end{gathered}$$

Thus, the lateral magnification is given using the above value in equation for |m2| as:

$$\begin{gathered} \left|m_2\right|=\frac{D-x_2}{x_2} \\ =\frac{{\displaystyle D-\frac{D}{2}-\frac{d}{2}}}{{\displaystyle \frac{D+d}{2}}} \end{gathered}$$

$=\frac{\frac{D-d}{2}}{{\displaystyle \frac{D+d}{2}}}$$=\frac{D-d}{D+d}$

The ratio of magnification is thus given as follows:

$$\begin{gathered} \frac{m_2}{m_1}=\frac{{\displaystyle \raisebox{1ex}{$\left(D-d\right)$}\!\left/ \!\raisebox{-1ex}{$(D+d)$}\right.}}{{\displaystyle \raisebox{1ex}{$(D+d)$}\!\left/ \!\raisebox{-1ex}{$(D-d)$}\right.}} \\ =\frac{{\left(D-d^{}\right)}^2}{{(D+d)}^2} \\ ={\left[\frac{D-d}{D+d}\right]}^2 \end{gathered}$$

Hence, the value of the required height ratio$={\left[\frac{D-d}{D+d}\right]}^2$