Question

You have a small but bright computer display that is only 4cm wide by 3cm tall. You will place a lens near the display to project an image of the display about 2.5m tall onto a large screen that is 6m from the lens of the projector. (a) Should you use a converging lens or a diverging lens? (b) Approximately, what should the focal length of the lens be? (c) Suppose that you choose a lens that has the focal length that you calculated in part (b). What exactly should be the distance from the computer display to the lens? (d) In order that the screen display be right side up, should the computer display be inverted or right side up? (e) On another occasion the screen is moved close, so that it is only 4m from the lens. To focus the image, you have to readjust the distance between the computer display and the lens. What is the new exact distance from the computer display to the lens? (f) How tall is the image on the screen, now that the screen is only 6m from the lens?

Short answer

a) Converging lens

b) The focal length of the lens should be 6.9cm.

c) The distance from the computer display to the lens should be 7cm.

d) In order that the screen display be right side up, the computer display be inverted.

e) The new exact distance from the computer display to the lens should be 7cm.

f) The image is 170cm tall on the screen

Step-by-step solution

Explain the given information

Consider a small but bright computer display that is 4cm wide and 3cm tall. A lens placed near the display that projects an image of the display about 2.5m tall onto a large screen that is 6m from the lens of the projector.

Give the Formula used.

Give the formula for magnification in terms of height and distance.

$\mathbf{\text{M=}}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{0}}}$ ……(1)

$\mathbf{\text{M=}}\frac{{\mathbf{\text{d}}}_{\mathbf{i}}}{{\mathbf{\text{d}}}_{\mathbf{0}}}$ ……(2)

Should we use a converging lens or a diverging lens 

a)

Considering the given setup, the image caught on the screen should be real. The real image can be obtained only by using the converging lens.

Therefore, a converging lens should be used for the given scenario.

Step 4: What should the focal length of the lens be?

b)

Consider the magnification formula in terms of height from Equation (1),

$\mathbf{\text{M=}}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{0}}}$

In the Equation (1), $h_1$is the height of the image and $h_0$ is the object distance.

Consider the magnification formula in terms of distance from Equation (2),

$\mathbf{\text{M=}}\frac{{\mathbf{\text{d}}}_{\mathbf{i}}}{{\mathbf{\text{d}}}_{\mathbf{0}}}$

In the Equation (2), $d_1$is the image distance and $d_0$is the object distance.

Compare the Equations (1) and (2),

$\frac{h_1}{h_0}=\frac{d_1}{d_0}$ ……(3)

Solve the Equation (3) for $d_0$ as follows,

$d_0=\frac{d_i{h}_0}{h_i}$ ……(4)

Substitute the values, $h_0=3\text{cm,}{d}_i=6\text{m,}{h}_i=2.5\text{m}$into Equation (4)

$\begin{array}{rcl}{d}_0& =& \frac{\left(600\text{cm}\right)\left(3\text{cm}\right)}{250\text{cm}}\\ d_0& =& 7.2\text{cm}\end{array}$ ……(5)

Consider the Equation of focal length,

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{i}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{0}}}$ ……(6)

Solve the Equation (6) for f as follows,

$f=\frac{d_i{d}_0}{d_i+d_0}$ ……(7)

Substitute the values, $d_i=600\text{cm}$ and $d_0=7.2\text{cm}$ into Equation (7)

$\begin{array}{rcl}f& =& \frac{\left(600\text{cm}\right)\left(7\text{cm}\right)}{600\text{cm}+\text{7cm}}\\ f& =& 6.9\text{cm}\end{array}$

Therefore, the focal length of the lens should be 6.9cm.

Step 5: What exactly should be the distance from the computer display to the lens?

c)

Consider the focal length equation as follows,

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{i}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{0}}}$ ……(6)

Solve the Equation (6) for $d_0$,

$d_0=\frac{f{d}_i}{d_i-f}$ ……(8)

Substitute the values $f=6.9\text{cm}$ and $d_i=600\text{cm}$ in the Equation (8),

$$\begin{gathered} d_0=\frac{\left(6.9\text{cm}\right)\left(600\text{cm}\right)}{600\text{cm}-\text{6.9cm}} \\ {d}_0=7\text{cm} \end{gathered}$$

Therefore, the distance from the computer display to the lens should be 7cm.

Step 6: In order that the screen display be right side up, should the computer display be inverted or right side up?

d)

Consider that the computer display is left side up, the image falls at right side down to get the image the right side up computer display should be left side down.

Hence, the computer display should be inverted to get image right side up.

Therefore, In order that the screen display be right side up, the computer display be inverted.

What is the new exact distance from the computer display to the lens?

e)

Consider the focal length equation as follows,

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{i}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{0}}}$ ……(6)

Solve the Equation (6) for $d_0$ ,

$d_0=\frac{f{d}_i}{d_i-f}$ ……(8)

Substitute the values $f=6.9\text{cm}$ and $d_i=400\text{cm}$ in the Equation (8),

$d_0=\frac{\left(6.9\text{cm}\right)\left(400\text{cm}\right)}{400\text{cm}-\text{6.9cm}}$

$d_0=7.02\text{cm}$

Therefore, the new exact distance from the computer display to the lens should be 7cm.

How tall is the image on the screen, now that the screen is only   from the lens?

f)

Consider the magnification formula in terms of height from Equation (1),

$\mathit{M}\mathbf{=}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{0}}}$

In the Equation (1), role="math" localid="1668575821721" $h_i$ is the height of the image and $h_0$ is the object distance.

Consider the magnification formula in terms of distance from Equation (2),

$\mathit{M}\mathbf{=}\frac{{\mathbf{d}}_{\mathbf{i}}}{{\mathbf{d}}_{\mathbf{0}}}$

In the Equation (2), $d_1$ is the image distance and $d_0$ is the object distance.

Compare the Equations (1) and (2),

$\frac{h_i}{h_0}=\frac{d_i}{d_0}$ ……(3)

Solve the Equation (3) for role="math" localid="1668575506465" $h_i$ as follows,

$h_i=\frac{d_i{h}_0}{d_0}$ ……(4)

Substitute the values, $h_0=3\text{cm,}{d}_i=400\text{cm,}{d}_0=7.02\text{cm}$ into Equation (4)

$$\begin{gathered} h_i=\frac{\left(400\text{cm}\right)\left(3\text{cm}\right)}{\text{7.02cm}} \\ {h}_i=170\text{cm} \end{gathered}$$ ……(5)

Therefore, the image is 170cm tall on the screen.