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Some converging light rays fall on a convex mirror of focal length of 30 cm. In absence of the mirror the rays could meet at distance of 20 cm from the pole of the mirror. Where do the rays meet in presence of the mirror?

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Aditya Bhutra ·2012-07-11 06:07:33

u= 20cm
f= 30 cm

1f = 1v - 1u

v= 12cm

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Sudipta Dhara ·2012-07-12 04:44:31

But the answere of the question is -60cm as given in the book.I have solved this after posting.I have taken u as +ve because rays could meet at that point.

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Aditya Bhutra ·2012-07-12 06:07:37

sorry , my mistake .
the negative sign in the formula should be changed to positive .

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Basic Integral Calculus Operators Symbols Matrix Cases

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∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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