JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
The position of the image formed by the combination of lenses is _______. \(\mathrm{f}_1=10 \mathrm{~cm} \quad \mathrm{f}_2=10 \quad \mathrm{f}_3=30 \mathrm{~cm}\)
- A \(30 \mathrm{~cm}\) (right of third lens)
- B \(15 \mathrm{~cm}\) (left of second lens)
- C \(30 \mathrm{~cm}\) (left of third lens)
- D \(15 \mathrm{~cm}\) (right of second lens)
Answer & Solution
Correct Answer
(A) \(30 \mathrm{~cm}\) (right of third lens)
Step-by-step Solution
Detailed explanation
For lens \(1: f_1=10, u=-30, v=\) ? \(v=\frac{u f}{u+f}=\frac{-30 \times 10}{-30+10}=15\) For lens 2: \(f_1=-10, u=10, v=\) ? \(\mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}+\mathrm{f}}=\frac{10 \times-10}{10-10}=\infty\) For lens \(3: f=30, u=-\infty, v=\) ? So v will be \(30\).
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