Important Derivations
Every derivation that appears in CBSE Class 10 Science board exams — step-by-step proofs for mirror formula, lens formula, resistances in series and parallel, and Joule's law.
Light — Reflection and Refraction
Mirror Formula
1/v + 1/u = 1/f
Steps
- 1
Draw a concave mirror with pole P, focus F, and centre of curvature C
- 2
Place object AB beyond C; draw two rays — one parallel to principal axis (reflects through F) and one through C (reflects back on itself)
- 3
Mark image A'B' at intersection of reflected rays
- 4
Using similar triangles ABP and A'B'P: A'B'/AB = PA'/PA → m = −v/u
- 5
Using similar triangles ABF and A'B'F: (PA−PF)/PF = PA'/PA → (u−f)/f = v/u
- 6
Rearrange: uv = uf + fv → divide both sides by uvf → 1/f = 1/v + 1/u
Use sign convention throughout: distances measured from pole. Object in front = negative u.
Magnification for Mirrors
m = −v/u = h'/h
Steps
- 1
Draw object AB and image A'B' for a concave mirror
- 2
Triangles ABP and A'B'P are similar (AA angle, right angle at base)
- 3
A'B'/AB = PA'/PA
- 4
With sign convention: h'/h = −v/u
- 5
Therefore m = h'/h = −v/u
m negative → inverted image (real). m positive → erect image (virtual). Must state this in answer.
Lens Formula
1/v − 1/u = 1/f
Steps
- 1
Draw a convex lens with optical centre O; place object AB on left side
- 2
Draw two rays: one parallel to principal axis (refracts through F₂) and one through optical centre (passes straight)
- 3
Mark image A'B' at intersection of refracted rays
- 4
Using similar triangles OAB and OA'B': A'B'/AB = OA'/OA → h'/h = v/u
- 5
Using similar triangles F₁OC and F₁A'B': A'B'/OC = (OA'−OF₁)/OF₁ → v/u = (v−f)/f
- 6
Cross multiply and divide by uvf: 1/f = 1/v − 1/u
Note: lens formula uses 1/v − 1/u = 1/f, NOT 1/v + 1/u like mirror formula.
Magnification for Lenses
m = v/u = h'/h
Steps
- 1
Triangles OAB and OA'B' are similar (vertically opposite angles at O)
- 2
A'B'/AB = OA'/OA
- 3
With sign convention: h'/h = v/u
- 4
Therefore m = v/u
For lenses, m = +v/u (no negative sign, unlike mirrors). Positive m = erect, negative m = inverted.
Electricity
Resistors in Series
R_s = R₁ + R₂ + R₃
Steps
- 1
In series, same current I flows through all resistors; voltage divides
- 2
V = V₁ + V₂ + V₃ (total voltage = sum of individual voltages)
- 3
By Ohm's law: V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
- 4
Therefore: IR_s = IR₁ + IR₂ + IR₃
- 5
Dividing by I: R_s = R₁ + R₂ + R₃
Key statement: 'In series combination, current is the same through all resistors.' Must write this.
Resistors in Parallel
1/R_p = 1/R₁ + 1/R₂ + 1/R₃
Steps
- 1
In parallel, same voltage V appears across each resistor; current divides
- 2
I = I₁ + I₂ + I₃ (total current = sum of branch currents)
- 3
By Ohm's law: I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃
- 4
Therefore: V/R_p = V/R₁ + V/R₂ + V/R₃
- 5
Dividing by V: 1/R_p = 1/R₁ + 1/R₂ + 1/R₃
Key statement: 'In parallel combination, voltage is the same across all resistors.' Equivalent resistance is always less than the smallest individual resistance.
Joule's Law of Heating
H = I²Rt
Steps
- 1
Work done to move charge Q through potential difference V: W = VQ
- 2
Since Q = It (charge = current × time): W = VIt
- 3
By Ohm's law, V = IR: W = IR × It = I²Rt
- 4
Heat produced H = work done W (assuming all electrical energy converts to heat)
- 5
Therefore H = I²Rt
- 6
Also written as H = VIt = V²t/R
Can also be derived as H = Pt where P = I²R. The three equivalent forms (H = I²Rt, VIt, V²t/R) may all be asked.
Electric Power
P = VI = I²R = V²/R
Steps
- 1
Power = work done per unit time: P = W/t
- 2
Work done in time t: W = VIt (from Joule's law derivation)
- 3
Therefore: P = VIt/t = VI
- 4
Substituting V = IR: P = (IR)I = I²R
- 5
Substituting I = V/R: P = V(V/R) = V²/R
All three forms are equivalent. Use whichever fits the given quantities. P = I²R used when current and resistance are given.