Geometrical Optics_JEE-Advanced-2022_S1 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE Advanced 2022

Paper-1 2022

Question

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.

Which one of the following options is correct?

Select the correct option:

(I) → P; (II) → R; (III) → Q; (IV) → T

(I) → Q; (II) → P; (III) → T; (IV) → S

(I) → P; (II) → T; (III) → R; (IV) → Q

(I) → T; (II) → S; (III) → Q; (IV) → R

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$ \begin{aligned} & (I) u=-20 \mathrm{~cm} \\ & f=+10 \mathrm{~cm} \\ & \frac{1}{v}+\frac{1}{20}=\frac{1}{10} \\ & \frac{1}{v}=\frac{1}{10}-\frac{1}{20} \\ & \frac{1}{v}=\frac{1}{20} \\ & v=20 \mathrm{~cm} \\ & u=+15 \mathrm{~cm} \\ & f=+15 \mathrm{~cm} \\ & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{15}=\frac{1}{15} \\ & \frac{1}{v}=\frac{2}{15} \\ & v=7.5 \mathrm{~cm}(\text { from lens } 2) \\ & I \rightarrow P \end{aligned} $
$ \begin{aligned} & \frac{1}{v}+\frac{1}{20}=\frac{1}{10} \\ & \frac{1}{v}=\frac{1}{10}-\frac{1}{20} \Rightarrow v=20 \mathrm{~cm} \end{aligned} $
$\begin{aligned} & u=+15 \mathrm{~cm} \\ & f=-10 \mathrm{~cm} \\ & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{15}=\frac{-1}{10} \\ & \frac{1}{v}=\frac{-1}{10}+\frac{1}{15}=\frac{-3+2}{30} \\ & \frac{1}{v}=-\frac{1}{30} \\ & v=-30 \mathrm{~cm} \\ & I I \rightarrow R \\ & (I I I) u=-20 \\ & f=+10 \mathrm{~cm} \\ & \frac{1}{v}+\frac{1}{20}=\frac{1}{10} \\ & \frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20} \\ & v=20 \mathrm{~cm} \\ & \Rightarrow u=15 \mathrm{~cm} \\ & f=-20 \mathrm{~cm}\end{aligned}$
$\begin{aligned} & \frac{1}{v}-\frac{1}{15}=\frac{11}{20} \\ & \frac{1}{v}=-\frac{1}{20}+\frac{1}{15}=\frac{-3+4}{60}=\frac{1}{60} \\ & v=60 \mathrm{~cm} \\ & \text { III } \rightarrow Q \\ & \text { (IV) } u=-20 \mathrm{~cm} \\ & f=-20 \\ & \Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \Rightarrow \frac{1}{v}+\frac{1}{20}=\frac{-1}{20} \\ & v=-10 \mathrm{~cm} \\ & u=-15 \mathrm{~cm} \\ & f=10 \mathrm{~cm} \\ & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \Rightarrow \frac{1}{v}+\frac{1}{15}=\frac{1}{10} \\ & \Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30} \\ & v=30 \mathrm{~cm} \\ & \text { IV } \rightarrow T\end{aligned}$

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JEE Advance

Physics

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