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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensMedium2 minQB

PhysicsMediumsingle choice

. A thin converging lens forms the image of an object at a distance of 30 cm on a screen at a distance 50 cm from the lens. The screen is then drawn closer by a distance of 1 mm. By what distance should the object be shifted so that its image on the screen is sharp again (approximately)?

Options:

Answer:

Solution:

$\frac{d v}{d u} = \frac{v ^{2}}{u ^{2}} = \frac{5 0 ^{2}}{3 0 ^{2}} = \frac{25}{9} d u = d v \times \frac{9}{25} = \frac{1 \times 9}{25} mm = 0.36 mm$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction at curved surface/Lens

⏱ 2mℹ️ Source: QB

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