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JEE Advance - Physics (2022 - Paper 1 Online - No. 18)

List I contains four combinations of two lenses (1 and 2) whose focal lengths (in $\mathrm{cm}$ ) are indicated in the figures. In all cases, the object is placed $20 \mathrm{~cm}$ from the first lens on the left, and the distance between the two lenses is $5 \mathrm{~cm}$. List II contains the positions of the final images.

List-I List-II
(I) JEE Advanced 2022 Paper 1 Online Physics - Geometrical Optics Question 17 English 1 (P) Final image is formed at $7.5 \mathrm{~cm}$ on the right side of lens 2 .
(II) JEE Advanced 2022 Paper 1 Online Physics - Geometrical Optics Question 17 English 2 (Q) Final image is formed at $60.0 \mathrm{~cm}$ on the right side of lens 2 .
(III) JEE Advanced 2022 Paper 1 Online Physics - Geometrical Optics Question 17 English 3 (R) Final image is formed at $30.0 \mathrm{~cm}$ on the left side of lens $2 .$
(IV) JEE Advanced 2022 Paper 1 Online Physics - Geometrical Optics Question 17 English 4 (S) Final image is formed at $6.0 \mathrm{~cm}$ on the right side of lens 2 .
(T) Final image is formed at $30.0 \mathrm{~cm}$ on the right side of lens 2 .

Which one of the following options is correct?

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

Explanation

(I) $$u = - 20$$ cm

$$f = + 10$$ cm

$${1 \over v} + {1 \over {20}} = {1 \over {10}}$$

$${1 \over v} = {1 \over {10}} - {1 \over {20}}$$

$${1 \over v} = {1 \over {20}}$$

$$v = 20$$ cm

$$u = + 15$$ cm

$$f = + 15$$ cm

$${1 \over v} - {1 \over u} = {1 \over f}$$

$$ \Rightarrow {1 \over v} - {1 \over {15}} = {1 \over {15}}$$

$${1 \over v} = {2 \over {15}}$$

$$v = 7.5$$ cm (from lens 2)

I $$\to$$ P

(II) $$u = - 20$$ cm, $$f = + 10$$ cm

$${1 \over v} + {1 \over {20}} = {1 \over {10}}$$

$${1 \over v} = {1 \over {10}} - {1 \over {20}} \Rightarrow v = 20$$ cm

$$u = + 15$$ cm

$$f = - 10$$ cm

$${1 \over v} - {1 \over u} = {1 \over f} \Rightarrow {1 \over v} - {1 \over {15}} = {{ - 1} \over {10}}$$

$${1 \over v} = {{ - 1} \over {10}} + {1 \over {15}} = {{ - 3 + 2} \over {30}}$$

$${1 \over v} = - {1 \over {30}}$$

$$v = - 30$$ cm

II $$\to$$ R

(III) $$u = - 20$$ cm

$$f = + 10$$ cm

$${1 \over v} + {1 \over {20}} = {1 \over {10}}$$

$$\frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$$

$$v = 20$$ cm

$$ \Rightarrow u = 15$$ cm

$$f = - 20$$ cm

$${1 \over v} - {1 \over {15}} = {{ - 1} \over {20}}$$

$${1 \over v} = - {1 \over {20}} + {1 \over {15}} = {{ - 3 + 4} \over {60}} = {1 \over {60}}$$

$$v = 60$$ cm

III $$\to$$ Q

(IV) $$u = - 20$$ cm

$$f = - 20$$ cm

$$ \Rightarrow {1 \over v} - {1 \over u} = {1 \over f}$$

$$ \Rightarrow {1 \over v} + {1 \over {20}} = {{ - 1} \over {20}}$$

$$v = - 10$$ cm

$$u = - 15$$ cm

$$f = 10$$ cm

$${1 \over v} - {1 \over u} = {1 \over f}$$

$${1 \over v} + {1 \over {15}} = {1 \over {10}}$$

$$ \Rightarrow {1 \over v} = {1 \over {10}} - {1 \over {15}}$$

$$ = {{3 - 2} \over {30}} = {1 \over {30}}$$

$$v = 30$$ cm

IV $$\to$$ T

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