Q105CP - University Physics with Modern Physics

Question

CALC (a) For a lens with focal length $\mathit{f}$, find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)?

Step-by-Step Solution

Answer

1. For a lens with focal length $f,$the smallest distance possible between the object and its real image is $4f$.
2. The graph of function of the distance of the object from lens is

Step 1: Define the focal length.

The focal length is the distance between the convex or concave mirror and the focal point of the mirror.

The relation between the distance of object s , the distance of the image $s\text{'}$and the focal length $f$ is 

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s\text{'}}$

Step 2: Determine the minimum distance

By the relation between the distance of object s , the distance of the image $s\text{'}$ and the focal length  $f$

The distance of image written as:

$s\text{'}=\frac{sf}{s-f}$

The distance between the object and image is

$$\begin{gathered} D=s\text{'}+s \\ =\frac{sf}{s-f}+s \\ =\frac{s^2}{s-f} \end{gathered}$$

The minimum distance between the object and image is change in distance with distance of object is zero.

$$\begin{gathered} \frac{dD}{ds}=0 \\ \frac{d}{ds}\left(\frac{s^2}{s-f}\right)=0 \\ \frac{2s}{s-f}-\frac{s^2}{{\left(s-f\right)}^2}=0 \\ s=2f \end{gathered}$$

Now, the distance of image is:

$$\begin{gathered} \frac{1}{2f}+\frac{1}{s\text{'}}=\frac{1}{f} \\ s\text{'}=2f \\ \therefore D_{min}=4f \end{gathered}$$

Hence, for a lens with focal length $f$ the smallest distance possible between the object and its real image is $4f$.

Step 3: Draw the graph.

The expression of distance between the object and image is $D=\frac{s^2}{s-f}$.

The graph between the minimum distance and distance of object is:

Hence, the graph of distance between the object and image is drawn above.