Thin lenses — lens equation, lens maker's equation: Worked examples

2.5.6 · D3 · Physics › Optics › Thin lenses — lens equation, lens maker's equation

Ye page ek "no surprises" drill hai. Hum parent note ke do boxed rules lete hain aur unhe har tarah ke input pe try karte hain: converging aur diverging lenses, objects near aur far, special focal-point cases, aur ek flat plate bhi. Agar tum ye sab kar sako, toh koi bhi exam version tumhe surprise nahi kar sakta.

Shuru karne se pehle, ek reminder un tools ka jo hum baar baar use karenge. Neeche sab kuch teen earned facts pe tika hua hai:

Scenario matrix

Har thin-lens problem ko socho jaise teen dials mein se ek-ek row choose kar rahe ho. Neeche ka matrix case-classes list karta hai; har worked example tagged hai us cell(s) se jahan wo fit hota hai.

Dial	Cases jo alag behave karte hain	Covered in
Lens type	converging $f > 0$ · diverging $f < 0$ · flat plate $f = \infty$	Ex 1,2,3,4,6 · Ex 5,7 · Ex 8
Object position (converging $f$ ke barabar)	beyond $2 f$ · at $2 f$ · between $f$ and $2 f$ · at $f$ (degenerate) · inside $f$ (virtual)	Ex 2 · Ex 3 · Ex 4 · Ex 6 · Ex 5(div)/Ex 7
Shape / sign of $R$	biconvex ( $+, -$ ) · plano-convex (one $\infty$ ) · biconcave ( $-, +$ ) · flat ( $\infty, \infty$ )	Ex 1 · Ex 8 · Ex 5 · Ex 8
Extra flavours	real-world word problem · exam-style twist (image height, magnifier)	Ex 6 (camera) · Ex 4,7

Wo ek "degenerate" input jisse sab darte hain — object at the focal point — wo Ex 6 hai. Wo ek "kuch nahi hota" input — flat plate — wo Ex 8 hai. Hum dono ko seedha handle karte hain.

Example 1 — Biconvex shape → focal length

(Cell: converging · biconvex $R_{1} > 0, R_{2} < 0$ )

Worked example

$f$ nikalo Ek glass lens ( $n = 1.5$ ) ki front surface $R_{1} = + 15 cm$ aur back surface $R_{2} = - 30 cm$ hai. Uski focal length nikalo.

Forecast: Dono faces bahar ki taraf bulge karte hain, toh ye converging hai — expect karo $f > 0$ , kahin do-teen tens of cm ke aas paas.

Step 1 — Har radius ka sign fix karo. Ye step kyun? Maker's equation ka koi matlab nahi jab tak har $R$ apna centre-of-curvature sign na carry kare. Front face ka centre daayein (outgoing) hai ⇒ $R_{1} = + 15$ . Back face ka centre bayein (incoming) hai ⇒ $R_{2} = - 30$ .

Step 2 — Maker's equation mein plug karo. Ye step kyun? Ye wahi ek rule hai jo shape ko $f$ mein convert karta hai. $\frac{1}{f} = (1.5 - 1) (\frac{1}{15} - \frac{1}{- 30}) = 0.5 (\frac{1}{15} + \frac{1}{30})$

Step 3 — Fractions add karo. Ye step kyun? Common denominator 30: $\frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$ . $\frac{1}{f} = 0.5 \cdot \frac{1}{10} = \frac{1}{20} cm^{- 1} \Rightarrow f = + 20 cm$

Verify: $f > 0$ ✔ (converging, jaise forecast tha). Units: $[cm^{- 1}]^{- 1} = cm$ ✔. Ek quick sanity check — ek symmetric biconvex $\pm 15$ deta $f = 15$ cm; back ko flatter ( $30$ ) banana chahiye $f$ ko lambaana, aur sach mein $20 > 15$ . ✔

Example 2 — Object beyond $2 f$

(Cell: converging · object beyond $2 f$ )

Ab hum objects ko har distance pe rakhte hain aur image ko march karte hue dekhte hain. Figure ko ek baar study karo — baaki ke saare converging examples bas is line pe ek point hain.

Figure — Thin lenses — lens equation, lens maker's equation

Example 3 — Object exactly at $2 f$

(Cell: converging · object at $2 f$ — symmetry point)

Example 4 — Object between $f$ and $2 f$ (exam twist: image height nikalo)

(Cell: converging · object between $f$ and $2 f$ · exam-style height twist)

Worked example Projector geometry

Wahi lens $f = + 20$ cm. Ek $2.0$ cm tall object $u = - 30$ cm pe rakha hai ( $f = 20$ aur $2 f = 40$ ke beech mein). $v$ , $m$ , aur image height $h^{'}$ nikalo.

Forecast: $f$ aur $2 f$ ke beech mein ⇒ image real, inverted, magnified, $2 f$ ke beyond form hogi. Ye exactly Magnification and image formation in action hai.

Step 1 — Lens equation. Ye step kyun? Pehle image locate karo; height ko $m$ chahiye, $m$ ko $v$ chahiye. $\frac{1}{v} = \frac{1}{20} + \frac{1}{- 30} = \frac{3 - 2}{60} = \frac{1}{60} \Rightarrow v = + 60 cm$

Step 2 — Magnification. $m = \frac{v}{u} = \frac{60}{- 30} = - 2$

Step 3 — Image height. Ye step kyun? $m = h^{'} / h$ , toh $h^{'} = m h$ . Ye twist hai: exam size chahta hai, sirf position nahi. $h^{'} = m h = (- 2) (2.0) = - 4.0 cm$

Verify: $v = 60 > 2 f = 40$ ✔ ( $2 f$ se beyond, jaise forecast tha). $∣ m ∣ = 2 > 1$ magnified ✔. $h^{'} = - 4.0$ cm: negative sign = inverted, aur $∣ h^{'} ∣ = 4.0 = 2 \times 2.0$ cm ✔. Height ki units cm hain ✔.

Example 5 — Diverging lens (biconcave)

(Cell: diverging $f < 0$ · biconcave $R_{1} < 0, R_{2} > 0$ · virtual image)

Worked example Ek problem mein do rules

Ek biconcave glass lens, $n = 1.5$ , $R_{1} = - 25 cm$ , $R_{2} = + 25 cm$ . Object $u = - 20$ cm pe khada hai. $f$ nikalo, phir $v$ aur $m$ .

Forecast: Concave ⇒ diverging ⇒ $f < 0$ . Ek diverging lens hamesha virtual, erect, diminished image deta hai ( $v < 0$ , $0 < m < 1$ ), chahe object kahin bhi ho.

Step 1 — $f$ ke liye maker's equation. Ye step kyun? Humein shape di gayi hai, $f$ nahi. Front face ka centre incoming (left) side pe hai ⇒ $R_{1} = - 25$ ; back face ka centre outgoing side pe hai ⇒ $R_{2} = + 25$ . $\frac{1}{f} = (0.5) (\frac{1}{- 25} - \frac{1}{+ 25}) = 0.5 (- \frac{2}{25}) = - \frac{1}{25} \Rightarrow f = - 25 cm$

Step 2 — $v$ ke liye lens equation. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 25} + \frac{1}{- 20} = \frac{- 4 - 5}{100} = - \frac{9}{100}$ $v = - \frac{100}{9} \approx - 11.11 cm$

Step 3 — Magnification. $m = \frac{v}{u} = \frac{- 100/9}{- 20} = \frac{100}{180} = \frac{5}{9} \approx + 0.556$

Verify: $f < 0$ ✔ (diverging). $v < 0$ ✔ (virtual, incoming side, figure mein dashed rays dekho). $0 < m < 1$ aur positive ✔ (erect, diminished). Sanity: virtual image lens ke paas hai object se ( $11.1 < 20$ ) — exactly wahi jo tumhare glasses karte hain jab tum near-sighted ho.

Example 6 — Object exactly at focal point (degenerate) + camera word problem

(Cell: converging · object AT $f$ — degenerate limit · real-world)

Worked example Jab

$u = - f$ ho toh image kahan jaati hai? (a) Converging lens $f = + 20$ cm, object exactly $u = - 20$ cm pe. $v$ nikalo. (b) Word problem: Ek camera mein $50$ mm ( $f = + 5.0$ cm) lens hai. Door ke pahaad ki photo kheenchne ke liye, sensor lens ke kitne peeche hona chahiye?

Forecast (a): Focal point pe object ka matlab hai rays lens ke baad parallel nikalti hain — wo kabhi reconverge nahi karti, toh image infinity pe hai ( $v \to \infty$ ). (b): Bahut door ka object doosri taraf se wahi same limit hai ( $u \to - \infty$ ): image seedha $f$ pe land honi chahiye.

Step 1 — Part (a), algebra aage badhao. Ye step kyun? Hum degeneracy equation mein dekhna chahte hain, sirf assert nahi karna. $\frac{1}{v} = \frac{1}{20} + \frac{1}{- 20} = 0 \Rightarrow v = \infty$ Reciprocal exactly zero hai, toh $v$ infinite hai: image kisi finite jagah form nahi hoti. Physically outgoing rays parallel hain.

Step 2 — Part (b), opposite limit. Ye step kyun? Ek pahaad effectively $u \to - \infty$ hai, toh $1/ u \to 0$ . $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{5.0} + 0 = \frac{1}{5.0} \Rightarrow v = + 5.0 cm$

Verify: (a) $1/ v = 0$ "image at infinity" confirm karta hai ✔ — isi liye $f$ pe rakha lit object spotlight/collimator banata hai. (b) $v = f = 5.0$ cm ✔: door ka object hamesha focal plane pe image banata hai, isliye sensor ek focal length peeche baithta hai. Dono parts wahi same degenerate case hain opposite sides se dekha hua. ✔

Example 7 — Magnifying glass (object inside $f$ )

(Cell: converging · object inside $f$ → virtual · exam twist: magnifier)

Worked example Ek converging lens virtual image bhi kaise de sakta hai

Converging lens $f = + 10$ cm. Ek stamp $u = - 6$ cm pe rakha hai (focal length ke andar). $v$ aur $m$ nikalo.

Forecast: $f$ ke andar, ek converging lens magnifying glass ki tarah kaam karta hai: virtual, erect, enlarged image ( $v < 0$ , $m > 1$ ).

Step 1 — Lens equation. Ye step kyun? Test karo ki "inside $f$ " sach mein $v$ ko negative flip karta hai ya nahi. $\frac{1}{v} = \frac{1}{10} + \frac{1}{- 6} = \frac{3 - 5}{30} = - \frac{2}{30} = - \frac{1}{15} \Rightarrow v = - 15 cm$

Step 2 — Magnification. $m = \frac{v}{u} = \frac{- 15}{- 6} = + 2.5$

Verify: $v = - 15 < 0$ ✔ (virtual, object ke same side — tum ise lens ke through dekhte ho). $m = + 2.5$ : positive ⇒ erect, $> 1$ ⇒ enlarged ✔. Note karo Ex 5 se contrast: wahan bhi negative $v$ tha, lekin yahan $∣ m ∣ > 1$ hai kyunki lens converge karta hai. Magnifier aur spectacles ka yahi poora farq hai. Dekho Power of a lens (dioptres).

Example 8 — Flat plate & near-flat limit

(Cell: flat plate $f = \infty$ · limiting behaviour · plano check)

Worked example "Kuch nahi karta" lens

(a) Ek flat slab of glass, $R_{1} = \infty$ , $R_{2} = \infty$ , $n = 1.5$ . $1/ f$ nikalo. (b) Ek plano-convex lens $R_{1} = \infty$ (flat front), $R_{2} = - 20$ cm, $n = 1.5$ . $f$ nikalo.

Forecast (a): Flat surfaces mein rays ko ek common point pe bend karne ki koi curvature nahi ⇒ $1/ f = 0$ , $f = \infty$ , koi focusing nahi. (b): Sirf ek face kaam karta hai, toh same $R$ ke biconvex se lamba $f$ expect karo.

Step 1 — Part (a): $R \to \infty$ limit lo. Ye step kyun? $1/ R \to 0$ jab $R \to \infty$ , toh har curvature term mar jaata hai. $\frac{1}{f} = (0.5) (\frac{1}{\infty} - \frac{1}{\infty}) = (0.5) (0 - 0) = 0 \Rightarrow f = \infty$ Ek flat plate optically inert hai (ye rays ko sideways shift karta hai lekin focus nahi karta).

Step 2 — Part (b): ek flat, ek curved. Ye step kyun? Dikhao ki ek dead surface simply drop out ho jaata hai. $\frac{1}{f} = (0.5) (\frac{1}{\infty} - \frac{1}{- 20}) = (0.5) (0 + \frac{1}{20}) = \frac{1}{40} \Rightarrow f = + 40 cm$

Verify: (a) $1/ f = 0$ ✔ — parent ke "flat plate kuch nahi karta" se match karta hai. (b) $f = + 40$ cm ✔, converging, aur biconvex $\pm 20$ se lamba (jo $20$ cm deta tha) kyunki sirf aadhi curvature kaam kar rahi hai ✔.

Recall

Recall Matrix ke across quick self-test

Converging lens ke $2 f$ se beyond object — image nature? ::: Real, inverted, diminished (Ex 2) Object at $2 f$ — magnification? ::: $m = - 1$ , real inverted same-size (Ex 3) Object at focal point — image kahan hai? ::: Infinity pe, $1/ v = 0$ (Ex 6a) Converging lens, object inside $f$ — image nature? ::: Virtual, erect, enlarged — magnifier (Ex 7) Diverging lens, koi bhi object — image nature? ::: Hamesha virtual, erect, diminished (Ex 5) Flat plate ( $R \to \infty$ ) ki focal length? ::: $f = \infty$ , $1/ f = 0$ , koi focusing nahi (Ex 8a)

Connections

Thin lenses — lens equation, lens maker's equation (parent)
Refraction at a single spherical surface
Snell's law & paraxial approximation
Magnification and image formation
Power of a lens (dioptres)
Lens combinations & equivalent focal length
Spherical mirrors — mirror equation
Lens aberrations (chromatic, spherical)