Given: object using the lens formula: ⇒⇒u′1⇒⇒u′1⇒u′f1=v′1−w′124021 cm=353 cm−u′1=353 cm−24021 cmu′1=51[73 cm−4821 cm]=51[336144−147]=−560 cm=−5.6 m Hence, the object should be at 5.6 m in front of the lens. Hence, the oblect should be moved by 5.6-24= 3.2 cm.
For an object placed at a distance 2.4m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass…
JEE Main 2022 — Physics Optics
For an object placed at a distance 2.4m from a lens, a sharp focused image is observed on a screen placed at a distance 12cm from the lens. A glass plate of refractive index 1.5 and thickness 1cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?
Held on 25 Jul 2022 · Verified 6 Jul 2026.
Options
- A
0.8m
- B
3.2m
- C
1.2m
- D
5.6m
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