A lens having focal length f and aperture of diameter d forms an image of intensity I . Aperture of diameter 2 d in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

Bycovering aperture, focal length does not change. But intensity is reduced by 4 1 times as aperture diameter 2 d is covered. ∴ I ′ = I − 4 I = 4 3 I ∴ New focal length = f and intensity = 4 3 I .

Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensMedium2 minPYQ_2016

PhysicsMediumsingle choice

A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$ . Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

Options:

Answer:

Solution:

Bycovering aperture, focal length does not change. But intensity is reduced by $\frac{1}{4}$ times as aperture diameter $\frac{d}{2}$ is covered. $∴ I^{'} = I - \frac{I}{4} = \frac{3 I}{4} ∴ New focal length = f and intensity = \frac{3 I}{4} .$

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

⏱ 2mℹ️ Source: PYQ_2016

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