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 respectivelyNEET 2010 (Screening)

Intensity, I ∝ A 2 ⇒ I 1 I 2 = [ A 1 A 2 ] 2 = π r 2 π r 2 − 4 π r 2 = 4 3 ⇒ I 2 = 4 3 I 1 and focal length remains unchanged.

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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsMiscellaneousEasy2 minPYQ_2010

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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 respectivelyNEET 2010 (Screening)

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Solution:

Intensity, $I \propto A^{2}$ $\Rightarrow \frac{I _{2}}{I _{1}} = [\frac{A _{2}}{A _{1}}]^{2} = \frac{π r ^{2} - \frac{π r ^{2}}{4}}{π r ^{2}} = \frac{3}{4}$ $\Rightarrow I_{2} = \frac{3}{4} I_{1}$ and focal length remains unchanged.

Stream:NEETSubject:PhysicsTopic:Wave OpticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2010

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