6.4 - Thin Lens Formula

Formula and Sign Convention

The thin lens formula connects object distance $u$ , image distance $v$ and focal length $f$ :

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}, \frac{1}{v} = \frac{1}{f} - \frac{1}{u}

Quantity	Sign convention
$f$	Positive for a convex/converging lens; negative for a concave/diverging lens.
$v$	Positive for a real image on the opposite side from the object; negative for a virtual image on the object’s side.

The illustrated real objects have positive $u$ .
With the same camera lens, a nearer object produces a larger image; a more distant object produces a smaller image.

Inference: Object distance influences image distance.
Hypothesis: Increasing object distance decreases image distance.
Aims: Study the relationship between $u$ and $v$ for a convex lens; determine focal length using the lens formula.
Variables: Manipulated — $u$ ; responding — $v$ ; constant — $f$ .
Apparatus: Convex lens with $f = 10.0 cm$ , lens holder, 6 V bulb on a wooden block, power supply, cardboard with a small triangular hole, white screen and metre rule.

Align the illuminated triangular hole, lens and screen along the metre rule. Carry out the experiment in a dark room.
Set $u = 15.0 cm$ . Adjust the screen until a sharp image forms.
Measure $v$ from the lens to the screen.
Repeat with $u = 20.0, 25.0, 30.0, 35.0 cm$ . Record $u$ , $v$ , $1/ u$ and $1/ v$ .
Plot $1/ v$ against $1/ u$ . Determine the gradient and both axis intercepts; use the lens formula to obtain $f$ .

Comparing $1/ v = - 1/ u + 1/ f$ with a straight-line equation gives a gradient of $- 1$ and intercepts of $1/ f$ ; the reciprocal of either intercept gives $f$ .

\frac{1}{v} = \frac{1}{12} - \frac{1}{18}, v = + 36 cm, m = \frac{36}{18} = 2

Image: real, inverted, magnified, 36 cm from the lens on the opposite side from the object.
$f = + 12 cm$ , $u = + 4 cm$ :

\frac{1}{v} = \frac{1}{12} - \frac{1}{4}, v = - 6 cm, m = \frac{6}{4} = 1.5

Image: virtual, upright, magnified, 6 cm from the lens on the object’s side. Use the magnitude of $v$ for the size ratio.

\frac{1}{v} = \frac{1}{- 30} - \frac{1}{60}, v = - 20 cm

\frac{h _{i}}{9} = \frac{20}{60}, h_{i} = 3 cm

Image: virtual, upright, diminished, 20 cm from the lens on the object’s side.
Cricket moving along the principal axis: For $f = 12 cm$ , moving from $u = 24 cm$ to $u = 18 cm$ changes image distance from 24 cm to 36 cm. Over 6 s, the image’s speed is:

\frac{36 - 24}{6} = 2 cm s^{- 1}