The thin lens equation 1f = 1d_o + 1d_i relates the focal... | SAT Equivalent expressions Question, Answer Explained

Math · Advanced MathHard

The thin lens equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ relates the focal length $f$ of a lens to the distance $d_o$ of an object from the lens and the distance $d_i$ of the object's image from the lens, where $d_o > f > 0$ . Which equation correctly expresses $d_i$ in terms of $f$ and $d_o$ ?

$d_i = d_o - f$

$d_i = \frac{d_o - f}{f d_o}$

$d_i = \frac{f d_o}{f - d_o}$

$d_i = \frac{f d_o}{d_o - f}$

Show answer and explanation

Answer: D

Subtract $\frac{1}{d_o}$ from both sides: $\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{d_o - f}{f d_o}$ . Take the reciprocal of both sides: $d_i = \frac{f d_o}{d_o - f}$ .

Choice B is the value of $\frac{1}{d_i}$ ; it still has to be flipped to give $d_i$ .

Why the other choices are wrong

A. Reciprocals don't subtract this way. The reciprocal of the image distance equals 1/f minus the reciprocal of the object distance.
B. This is the reciprocal of the image distance. Flip it to get the image distance itself.
C. This reverses the subtraction. Combining 1/f minus 1 over the object distance gives a numerator of the object distance minus f, not f minus the object distance.

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