Q106P - University Physics with Modern Physics

Question

An Object at an Angle. A 16.0cm  long pencil is placed at a 45.0° angle, with its center 15.0cm above the optic axis and 45.0 cm from a lens with a 20.0cm  focal length as shown in Fig. P34.106. (Note that the figure is not drawn to scale.) Assume that the diameter of the lens is large enough for the paraxial approximation to be valid. (a) Where is the image of the pencil? (Give the location of the images of the points and on the object, which are located at the eraser, point, and center of the pencil, respectively.) (b) What is the length of the image (that is, the distance between the images of points and )? (c) Show the orientation of the image in a sketch.

 

Step-by-Step Solution

Answer

1.   The location of images of the points A,B and C are 33.0 cm , 40.7 cm and 36.0 cm .
2. The length of the image is 16.46 cm.
3. The orientation of the image of the pencil is shown in the diagram below. 

Step 1: Given Data

Length of the pencil:  16.0 cm

Inclination of pencil:  $45°$

Distance of midpoint from the axis:  15.0 cm

Distance of midpoint from the lens:  45.0 cm

Focal length: 20.0 cm

Step 2: Define the focal length.

The focal length of a lens is the perpendicular distance of the focal plane from the optical center of the lens, along the principal axis.

The relation between the distance of object s , the distance of the image, and the focal length  $f$ is

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

Step 3: Determine the location of image.

The distance of Image of points A,B and C from the lens are:

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

For point the focal length $f= 20 \mathrm{cm}$and distance of object s is $45 \mathrm{cm} +\left(8 \mathrm{cm}\right)\cos 45°=50.7 \mathrm{cm}$.

So, 

$$\begin{gathered} s_A^{\text{'}}=\frac{\left(50.7 \mathrm{cm}\right)\left(20 \mathrm{cm}\right)}{50.7 \mathrm{cm}-20 \mathrm{cm}} \\ =33.0 \mathrm{cm} \end{gathered}$$

For point $B$ the focal length $f=20 \mathrm{cm}$and distance of object  s is $45 cm-\left(8 cm\right)\cos 45°=39.3 cm.$

So, $$\begin{gathered} s_B^{\text{'}}=\frac{\left(39.3 \mathrm{cm}\right)\left(20 \mathrm{cm}\right)}{39.3 \mathrm{cm}-20 \mathrm{cm}} \\ =40.7 \mathrm{cm} \end{gathered}$$

For point B the focal length $f= 20 cm$ and distance of object $s$ is 45.0 cm.

So,$$\begin{gathered} s_C^{\text{'}}=\frac{\left(45 \mathrm{cm}\right)\left(20 \mathrm{cm}\right)}{45 \mathrm{cm}-20 \mathrm{cm}} \\ =36.0 \mathrm{cm} \end{gathered}$$

Hence, the location of images of the points A,B and C are 33.0cm, 40.7 cm and 36.0 cm

Step 4: Determine the length of image.

The height of image at points A and B are:

$$\begin{gathered} h_A^{\text{'}}=\frac{s\text{'}}{s}{h}_A \\ =\frac{33 \mathrm{cm}}{45 \mathrm{cm}}\left(15 \mathrm{cm}-8 \mathrm{cm}\left(\sin 45°\right)\right) \\ =-6.85 \mathrm{cm} \\ {h}_B^{\text{'}}=\frac{s\text{'}}{s}{h}_B \\ =\frac{40.7 \mathrm{cm}}{39.3 \mathrm{cm}}\left(15 \mathrm{cm}-8 \mathrm{cm}\left(\sin 45°\right)\right) \\ =-21.4 \mathrm{cm} \end{gathered}$$

Now, length of image is:

$$\begin{gathered} L=\sqrt{{\left({\mathrm{s}}_{\mathrm{A}}^{\text{'}}-{\mathrm{s}}_{\mathrm{B}}^{\text{'}}\right)}^2{\left({\mathrm{h}}_{\mathrm{A}}-{\mathrm{h}}_{\mathrm{B}}\right)}^2} \\ =\sqrt{{\left(33 \mathrm{cm}-40.7 \mathrm{cm}\right)}^2{\left(6.87 \mathrm{cm}-21.4 \mathrm{cm}\right)}^2} \\ =16.46 \mathrm{cm} \end{gathered}$$

Hence, the length of the image is $=16.46 \mathrm{cm}$.

Step 5: Determine orientation of the image.

The orientation of the image of the pencil is shown on the right side of the lens.