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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2024
  • Hard
  • MCQ

The position of the image formed by the combination of lenses is _______. \(\mathrm{f}_1=10 \mathrm{~cm} \quad \mathrm{f}_2=10 \quad \mathrm{f}_3=30 \mathrm{~cm}\)

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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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