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The owner of a van installs a rear-window lens that has a focal length of -0.340 m. When the owner looks out through the lens at an object located directly behind the van, the object appears to be 0.240 m from the back of the van, and appears to be 0.390 m tall. (a) How far from the van is the object actually located, and (b) how tall is the object?

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Answer:

0.816 m

1.326 m

Explanation:

u = Object distance

v = Image distance = -0.24

f = Focal length = -3 m (concave lens)

[tex]h_v[/tex] = Image height = 0.39 m

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{-0.34}=\frac{1}{u}+\frac{1}{-0.24}\\\Rightarrow \frac{1}{u}=\dfrac{1}{-0.34}+\dfrac{1}{0.24}\\\Rightarrow u=0.816\ m[/tex]

The object is 0.816 m from the van

[tex]m=\frac{h_v}{h_u}\\\Rightarrow -\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow h_u=-\dfrac{h_vu}{v}\\\Rightarrow h_u=-\dfrac{0.39\times 0.816}{-0.24}\\\Rightarrow h_u=1.326\ m[/tex]

The height of the object is 1.326 m

okpalawalter8

A) The object is located at a distance of

  • 0.816m from the van

Using lens equation,

[tex]\frac{1}{u}+\frac{1}{v} = \frac{1}{f}\\\\\frac{1}{u} = \frac{1}{f} - \frac{1}{v}\\\\\frac{1}{u} = \frac{1}{-0.34} - \frac{1}{-0.24}\\\\u = 0.816m[/tex]

B) The object is

  • 0.936m tall

[tex]\frac{h_i}{h_o} = \frac{-v}{u}\\\\\frac{0.39}{h_o} = \frac{-(-0.34)}{0.816}\\\\h_o = 0.936m[/tex]

What is focal length?

Focal length, usually represented in millimeters (mm), is the basic description of a photographic lens. Lens focal length tells us the angle of view, how much of the scene will be captured, and the magnification, how large individual elements will be.

For more information on focal length, visit

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