Equivalent Power of Two Lenses at Distance

 

❓ Question

How do we calculate the equivalent power when two thin lenses are placed at a distance from each other?

---

🖼 Concept Image

---

✍️ Short Concept

This is a lens combination formula-based question.

Golden idea:

👉 Touching lenses → simple addition
👉 Separated lenses → correction term aata hai

---

🔷 Step 1 — Power of a Lens 💯

$$P = \frac{1}{f}$$

(f in metres)

Convex lens → positive power
Concave lens → negative power

---

🔷 Step 2 — When Lenses Are Touching

If lenses are in contact:

$$\boxed{P_{total} = P_1 + P_2}$$

👉 Bas simple addition.
No extra term.

---

🔷 Step 3 — When Lenses Are Separated (MOST IMPORTANT)

If separation distance = d:

$$\boxed{P_{total} = P_1 + P_2 - d P_1 P_2}$$

⚠️ Distance term always minus
⚠️ d must be in metres

Yahin JEE trap lagata hai.

---

🔷 Step 4 — Why Distance Matters?

First lens image banata hai.

Wahi image second lens ke liye object ban jata hai.

Isliye system simple addition jaisa behave nahi karta.

Mathematically → extra correction term add hota hai.

---

🔷 Step 5 — JEE Golden Mistakes

❌ Direct power add kar dena
❌ cm ko metre mein convert na karna
❌ Minus sign bhool jaana

👉 0.1 m vs 10 cm = huge difference.

---

✅ Final Takeaway

Touching lenses:

$$P_{eq} = P_1 + P_2$$

Separated lenses:

$$P_{eq} = P_1 + P_2 - dP_1P_2$$

Bas ye do lines yaad.

---

⭐ Golden JEE Insight

If:

$$d = 0$$

Formula automatically becomes:

$$P_{eq} = P_1 + P_2$$

🔥 Means touching case is special case of general formula.