JEE Main 2026 — 5 April, Evening Shift — Question 15

For lens P: $1/v - 1/(-15) = 1/10$ ⇒ $v=+30$ cm (30 cm right of P, so 15 cm left of Q). For lens Q: object distance = +15 cm (virtual object? Actually object for Q is 15 cm to its left, so u = +15 cm? Sign convention: for Q, object is on left, so u = -15 cm? Let's follow: $1/v' - 1/u = 1/f$, with u = -15 cm, f=+15 cm ⇒ $1/v' = 1/15 + 1/(-15) = 0$ ⇒ v' = ∞? That gives infinity. But solution says V' = 7.5 cm. There's confusion. The given solution uses $1/V' - 1/15 = 1/15$ ⇒ V' = 7.5 cm, meaning they took u = +15 cm (object to the right? Actually after first lens, image is at 30 cm right of P, which is 15 cm right of Q? No: Q is 15 cm right of P, so image from P is at 30 cm right of P = 15 cm right of Q. That is a real image on the right side of Q, so for Q it acts as a virtual object? The solution treats it as u = +15 cm (object on the right). Then magnification = (30/-15)*(7.5/15) = -1, same size, real image.