Verified Solution • 2025–26 Session moderate

QJEEAdv22Ph4d86: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph4d86 Solution: List I contains four combinations of two lenses (1 and 2) whose …

List I contains four combinations of two lenses (1 and 2) whose focal lengths (in cm\mathrm{cm} ) are indicated in the figures. In all cases, the object is placed 20 cm20 \mathrm{~cm} from the first lens on the left, and the distance between the two lenses is 5 cm5 \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 cm7.5 \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 cm60.0 \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 cm30.0 \mathrm{~cm} on the left side of lens 2.2 .
(IV) JEE Advanced 2022 Paper 1 Online Physics - Geometrical Optics Question 17 English 4 (S) Final image is formed at 6.0 cm6.0 \mathrm{~cm} on the right side of lens 2 .
(T) Final image is formed at 30.0 cm30.0 \mathrm{~cm} on the right side of lens 2 .

Which one of the following options is correct?

Multiple Choice Options

  • A (I) →\rightarrow P; (II) →\rightarrow R; (III) →\rightarrow Q; (IV) →\rightarrow T
  • B (I) →\rightarrow Q; (II) →\rightarrow P; (III) →\rightarrow T; (IV) →\rightarrow S
  • C (I) →P\rightarrow \mathrm{P} ; (II) →T\rightarrow \mathrm{T} ; (III) →R\rightarrow \mathrm{R} ; (IV) →Q\rightarrow \mathrm{Q}
  • D (I) →\rightarrow T; (II) →\rightarrow S; (III) →\rightarrow Q; (IV) →\rightarrow R

Step-by-Step Verified Solution

(I) u=−20u = - 20 cm

f=+10f = + 10 cm

1v+120=110{1 \over v} + {1 \over {20}} = {1 \over {10}}

1v=110−120{1 \over v} = {1 \over {10}} - {1 \over {20}}

1v=120{1 \over v} = {1 \over {20}}

v=20v = 20 cm

u=+15u = + 15 cm

f=+15f = + 15 cm

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

⇒1v−115=115\Rightarrow {1 \over v} - {1 \over {15}} = {1 \over {15}}

1v=215{1 \over v} = {2 \over {15}}

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

I →\to P

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

1v+120=110{1 \over v} + {1 \over {20}} = {1 \over {10}}

1v=110−120⇒v=20{1 \over v} = {1 \over {10}} - {1 \over {20}} \Rightarrow v = 20 cm

u=+15u = + 15 cm

f=−10f = - 10 cm

1v−1u=1f⇒1v−115=−110{1 \over v} - {1 \over u} = {1 \over f} \Rightarrow {1 \over v} - {1 \over {15}} = {{ - 1} \over {10}}

1v=−110+115=−3+230{1 \over v} = {{ - 1} \over {10}} + {1 \over {15}} = {{ - 3 + 2} \over {30}}

1v=−130{1 \over v} = - {1 \over {30}}

v=−30v = - 30 cm

II →\to R

(III) u=−20u = - 20 cm

f=+10f = + 10 cm

1v+120=110{1 \over v} + {1 \over {20}} = {1 \over {10}}

1v=110−120=120\frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}

v=20v = 20 cm

⇒u=15\Rightarrow u = 15 cm

f=−20f = - 20 cm

1v−115=−120{1 \over v} - {1 \over {15}} = {{ - 1} \over {20}}

1v=−120+115=−3+460=160{1 \over v} = - {1 \over {20}} + {1 \over {15}} = {{ - 3 + 4} \over {60}} = {1 \over {60}}

v=60v = 60 cm

III →\to Q

(IV) u=−20u = - 20 cm

f=−20f = - 20 cm

⇒1v−1u=1f\Rightarrow {1 \over v} - {1 \over u} = {1 \over f}

⇒1v+120=−120\Rightarrow {1 \over v} + {1 \over {20}} = {{ - 1} \over {20}}

v=−10v = - 10 cm

u=−15u = - 15 cm

f=10f = 10 cm

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

1v+115=110{1 \over v} + {1 \over {15}} = {1 \over {10}}

⇒1v=110−115\Rightarrow {1 \over v} = {1 \over {10}} - {1 \over {15}}

=3−230=130= {{3 - 2} \over {30}} = {1 \over {30}}

v=30v = 30 cm

IV →\to T

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