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Physics · Geometrical Optics

JEE Advanced 2022 — Paper 1 — Question 34

List I contains four combinations of two lenses (1 and 2) whose focal lengths (in cm ) are indicated in the figures.

In all cases, the object is placed 20 cm from the first lens on the left, and the distance between the two lenses is 5 cm .

List II contains the positions of the final images.

List -IList -II
(I)figure(P)Final image is formed at 7.5 cm on the right side of lens 2 .
(II)figure(Q)Final image is formed at 60.0 cm on the right side of lens 2 .
(III)figure(R)Final image is formed at 30.0 cm on the left side of lens 2.
(IV)figure(S)Final image is formed at 6.0 cm on the right side of lens 2 .
(T)Final image is formed at 30.0 cm on the right side of lens 2.
  1. Option A:

    (I) →P\rightarrow \mathrm{P}; (II) →R\rightarrow \mathrm{R}; (III) →Q\rightarrow \mathrm{Q}; (IV) →T\rightarrow \mathrm{T}

    Correct
  2. Option B:

    (I) →Q\rightarrow \mathrm{Q}; (II) →P\rightarrow \mathrm{P}; (III) →T\rightarrow \mathrm{T}; (IV) →S\rightarrow \mathrm{S}

  3. Option C:

    (I) →P\rightarrow \mathrm{P}; (II) →T\rightarrow \mathrm{T}; (III) →R\rightarrow \mathrm{R}; (IV) →Q\rightarrow \mathrm{Q}

  4. Option D:

    (I)→T(\mathrm{I}) \rightarrow \mathrm{T}; (II) →S\rightarrow \mathrm{S};(III) →Q;(IV)→R\rightarrow \mathrm{Q} ;(\mathrm{IV}) \rightarrow \mathrm{R}

Answer: A

Step-by-step solution

1v=1f+1u\frac{1}{v}=\frac{1}{f}+\frac{1}{u} (concavelens)

⇒v=−10\Rightarrow \mathrm{v}=-10

1v=1f+1u(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}( convex lens ))

1v=110+1−15\frac{1}{v}=\frac{1}{10}+\frac{1}{-15}

⇒v=+30\Rightarrow \mathrm{v}=+30

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Physics
Chapter
Geometrical Optics
Topic
Lenses and Their Combinations, Silvering of Lens

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