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The thin-lens formula relating focal length ff, object distance dod_o, and image distance did_i is given by 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}. Rather than isolating the focal length, solve this equation explicitly for the object distance dod_o in terms of ff and did_i, expressing the final result as a single simplified rational expression.

Answer and steps

Final answer

do=fdidi−fd_o = \frac{f d_i}{d_i - f}

Steps

01

Isolate the term containing dod_o

To solve the thin-lens formula for the object distance dod_o, isolate the term containing dod_o and then take the reciprocal.

Start with the given equation: 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

Subtract 1di\frac{1}{d_i} from both sides to get 1do\frac{1}{d_o} by itself: 1do=1f−1di\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}

02

Combine the fractions on the right side

To combine the terms on the right-hand side, find the common denominator, which is fdif d_i: 1f=difdi\frac{1}{f} = \frac{d_i}{f d_i} 1di=ffdi\frac{1}{d_i} = \frac{f}{f d_i}

Substitute these back into the equation: 1do=di−ffdi\frac{1}{d_o} = \frac{d_i - f}{f d_i}

03

Solve for dod_o by taking the reciprocal

Take the reciprocal of both sides of the equation: do=fdidi−fd_o = \frac{f d_i}{d_i - f}

Common Pitfall to Avoid

A frequent mistake when working with reciprocal equations is attempting to invert each term individually: 1do=1f−1di  ⟹  do≠f−di\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i} \implies d_o \neq f - d_i Taking the reciprocal is an operation applied to the entire side of the equation. You must first combine the terms into a single rational expression before inverting both sides.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student reciprocates individual terms directly across addition or subtraction instead of reciprocating the entire combined expression.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    1do=1f−1di\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}

  2. Step 2

    d_o = f - d_i

    Correct

    1do\frac{1}{d_o} = di−ffdi\frac{d_i - f}{f d_i}   ⟹  \implies d_o = fdidi−f\frac{f d_i}{d_i - f}

    You took the reciprocal of each individual term on the right-hand side. Because 1a−b≠1a−1b\frac{1}{a - b} \ne \frac{1}{a} - \frac{1}{b}, you must combine the terms over a common denominator first, giving di−ffdi\frac{d_i - f}{f d_i}, and then invert the entire fraction.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly isolated the target variable's term 1do\frac{1}{d_o} on one side of the equation.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Term-by-term reciprocal error

You treated reciprocal inversion as distributing over subtraction, writing the reciprocal of 1f−1di\frac{1}{f} - \frac{1}{d_i} as f−dif - d_i.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the reciprocal of a difference with the difference of reciprocals. Taking the reciprocal is an operation applied to an entire side of an equation, not to each term separately.

REASONING PATH
Your ACEpath teacher’s suggested approach
Combine before inverting

  • The reciprocal of a sum or difference is 1A±B\frac{1}{A \pm B}, which is never equal to 1A±1B\frac{1}{A} \pm \frac{1}{B}.

  • Find a common denominator to merge multiple fractions into a single fraction PQ\frac{P}{Q} before flipping to QP\frac{Q}{P}.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Write the right-hand side with a common denominator of fdif d_i: 1do=difdi−ffdi\frac{1}{d_o} = \frac{d_i}{f d_i} - \frac{f}{f d_i}

  2. Combine into a single fraction: 1do=di−ffdi\frac{1}{d_o} = \frac{d_i - f}{f d_i}

  3. Invert both sides to solve for dod_o: do=fdidi−fd_o = \frac{f d_i}{d_i - f}

Final answer

d_o = fdidi−f\frac{f d_i}{d_i - f}

NEXT STEPS
Final Takeaways from Your Teacher

  • Always combine separate fraction terms into one fraction before flipping.

  • Test your result with numbers (for example, if f=2f=2 and di=3d_i=3, check whether 12−13=16\frac{1}{2} - \frac{1}{3} = \frac{1}{6} gives do=6d_o = 6, rather than 2−3=−12 - 3 = -1).

OverviewACEpath Teacher's Diagnosis

02 The student subtracts terms incorrectly when finding a common denominator, reversing the terms in the numerator.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    1do=1f−1di\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}

  2. Step 2

    1do=f−dif⋅di\frac{1}{d_o} = \frac{f - d_i}{f \cdot d_i}

    Correct

    1do=di−ffdi\frac{1}{d_o} = \frac{d_i - f}{f d_i}

    When finding a common denominator for 1f−1di\frac{1}{f} - \frac{1}{d_i}, the first term becomes difdi\frac{d_i}{f d_i} and the second becomes ffdi\frac{f}{f d_i}, giving di−ffdi\frac{d_i - f}{f d_i} rather than f−difdi\frac{f - d_i}{f d_i}.

  3. Step 3

    d_o = f⋅dif−di\frac{f \cdot d_i}{f - d_i}

    Correct

    d_o = fdidi−f\frac{f d_i}{d_i - f}

    You correctly inverted both sides of your equation, but the final expression carries over the sign error from the previous numerator.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly isolated the term containing the target variable on one side of the equation.

You recognized that after combining the right-hand side into a single fraction, taking the reciprocal yields the isolated variable.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Order of terms in subtraction during common denominator conversion

You reversed the order of the numerator terms when subtracting the fractions, writing f−dif - d_i instead of di−fd_i - f.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Losing track of which term corresponds to which fraction during cross-multiplication or common denominator expansion. Because subtraction is not commutative, 1a−1b=b−aab\frac{1}{a} - \frac{1}{b} = \frac{b - a}{ab}, not a−bab\frac{a - b}{ab}.

REASONING PATH
Your ACEpath teacher’s suggested approach
Track the Minuend and Subtrahend

  • Subtraction is order-dependent: 1a−1b=bab−aab=b−aab\frac{1}{a} - \frac{1}{b} = \frac{b}{ab} - \frac{a}{ab} = \frac{b - a}{ab}.

  • Write out the intermediate step 1⋅difdi−1⋅ffdi\frac{1 \cdot d_i}{f d_i} - \frac{1 \cdot f}{f d_i} before combining into a single fraction.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Write each fraction with the common denominator fdif d_i: 1do=difdi−ffdi\frac{1}{d_o} = \frac{d_i}{f d_i} - \frac{f}{f d_i}.

  2. Combine the numerators in order: 1do=di−ffdi\frac{1}{d_o} = \frac{d_i - f}{f d_i}.

  3. Take the reciprocal of both sides: do=fdidi−fd_o = \frac{f d_i}{d_i - f}.

Final answer

d_o = fdidi−f\frac{f d_i}{d_i - f}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice rewriting subtractions of unit fractions like 1x−1y\frac{1}{x} - \frac{1}{y} step-by-step to reinforce numerator order.

  • Always double-check by testing numerical values: if f=2f = 2 and di=3d_i = 3, verify that 12−13=3−26=16>0\frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} > 0.

OverviewACEpath Teacher's Diagnosis

03 The student adds the fractions instead of subtracting after moving 1/d_i to the other side.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    1do=1f+1di\frac{1}{d_o} = \frac{1}{f} + \frac{1}{d_i}

    Correct

    1do=1f−1di\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}

    To isolate 1do\frac{1}{d_o} from the equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, you must apply the subtraction property of equality and subtract 1di\frac{1}{d_i} from both sides. Instead, the term was added.

  2. Step 2

    1do=di+ff⋅di\frac{1}{d_o} = \frac{d_i + f}{f \cdot d_i}

    Correct

    1do=di−ffdi\frac{1}{d_o} = \frac{d_i - f}{f d_i}

    Your fraction addition using the common denominator fdif d_i was algebraically valid for your equation, but it inherited the incorrect plus sign from Step 1.

  3. Step 3

    d_o = f⋅didi+f\frac{f \cdot d_i}{d_i + f}

    Correct

    d_o = fdidi−f\frac{f d_i}{d_i - f}

    You correctly took the reciprocal of both sides, but the final expression carries forward the sign error from the initial step.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified that finding a single rational expression on the right-hand side is required before inverting.

You accurately found the common algebraic denominator f⋅dif \cdot d_i.

You correctly inverted both sides of the equation to isolate dod_o.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign Error in Transposition

You wrote a plus sign instead of a minus sign when moving 1di\frac{1}{d_i} across the equals sign to isolate 1do\frac{1}{d_o}.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Overlooking the subtraction property of equality when transposing an added term across an equals sign.

REASONING PATH
Your ACEpath teacher’s suggested approach
Additive Inverse Check

  • To move an added term +B+B to the other side of an equation A=X+BA = X + B, you must perform the inverse operation: subtract BB from both sides (X=A−BX = A - B).

  • Always write out −1di-\frac{1}{d_i} on both sides explicitly before simplifying.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Start with 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} and subtract 1di\frac{1}{d_i} from both sides: 1do=1f−1di\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}.

  2. Combine terms using the common denominator fdif d_i: 1do=di−ffdi\frac{1}{d_o} = \frac{d_i - f}{f d_i}.

  3. Invert both sides to solve for dod_o: do=fdidi−fd_o = \frac{f d_i}{d_i - f}.

Final answer

d_o = fdidi−f\frac{f d_i}{d_i - f}

NEXT STEPS
Final Takeaways from Your Teacher

  • Double-check operation signs whenever transposing terms across the equals sign.

  • Practice isolating variables in formulas with two or more reciprocal fractions.

Lock it in

Question

Does 1x+1y=1z\frac{1}{x} + \frac{1}{y} = \frac{1}{z} mean that x+y=zx + y = z?

Answer

No. Taking the reciprocal applies to the whole side, not term-by-term: 11x+1y=xyx+y≠x+y\frac{1}{\frac{1}{x} + \frac{1}{y}} = \frac{xy}{x+y} \ne x+y.

Similar question

01

Solve 1Rp=1R1+1R2\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} explicitly for R1R_1 in terms of RpR_p and R2R_2.

Answer

R1=RpR2R2−RpR_1 = \frac{R_p R_2}{R_2 - R_p}. Subtract 1R2\frac{1}{R_2} to get 1R1=1Rp−1R2\frac{1}{R_1} = \frac{1}{R_p} - \frac{1}{R_2}, combine into R2−RpRpR2\frac{R_2 - R_p}{R_p R_2}, then take the reciprocal.

02

Solve 2a=1b−3c\frac{2}{a} = \frac{1}{b} - \frac{3}{c} for aa in terms of bb and cc, writing the result as a single simplified fraction.

Answer

a=2bcc−3ba = \frac{2bc}{c - 3b}. Combine the right side: 1b−3c=c−3bbc\frac{1}{b} - \frac{3}{c} = \frac{c - 3b}{bc}. Invert both sides: a2=bcc−3b\frac{a}{2} = \frac{bc}{c - 3b}, then multiply by 22.

03

Solve 1x+1+1y=1k\frac{1}{x+1} + \frac{1}{y} = \frac{1}{k} for xx in terms of yy and kk.

Answer

x=ky−y+ky−kx = \frac{ky - y + k}{y - k} (or x=kyy−k−1x = \frac{ky}{y - k} - 1). Isolate 1x+1=1k−1y=y−kky\frac{1}{x+1} = \frac{1}{k} - \frac{1}{y} = \frac{y - k}{ky}, take the reciprocal to get x+1=kyy−kx + 1 = \frac{ky}{y - k}, and subtract 11.

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