doubt in optics

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A diverging lens is placed at origin of coordinate sysytem.eqn of incident ray is y=-x/2+3. The focal length of lens is 4 cm. find eqn of divergent ray...
please help....anyone..

#Physics #General Physics

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6 Answers

govind ·2010-03-16 10:50:55

This question has been dicussed earlier here...
http://www.targetiit.com/iit-jee-forum/posts/a-question-on-convex-lens-14509.html

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var ·2010-03-17 01:01:05

can anyone give a detailed solution...

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var ·2010-03-18 02:53:22

???..

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Smriti Kumari ·2010-03-18 10:09:34

its y-3=x/4.
u=6 (ray wud hv intersected x-axis there)
so, v=-12
2 pts for the refracted ray are (-12,0) & (0,3)

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var ·2010-03-19 03:44:42

tthough d ans is correct...i didnt get d u=6 part...isnt d ray coming from infinity......

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eureka123 ·2010-03-19 22:22:08

@smriti....how did u get that eqn ??

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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

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

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

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

° ∂ ∞ 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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doubt in optics

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