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Geometrical Optics question

2010 · Q151
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Geometrical Optics question

2010 · Q151

NEETPhysicsGeometrical OpticsMCQ+4 / −1
A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d2{d \over 2}2d​ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively
  1. A
    f f\,f and I4{I \over 4}4I​
  2. B
    3f4{{3f} \over 4}43f​ and I2{I \over 2}2I​
  3. C
    fff and 3I4{{3I} \over 4}43I​
  4. D
    f2{f \over 2}2f​ and I2{I \over 2}2I​
View written solutionFree

Correct answer: C

By covering aperture, focal length does not change. But intensity is reduced by 14{1 \over 4}41​ times, as aperture diameter d2{d \over 2}2d​ is covered.

∴\therefore∴ I' = I - I4{I \over 4}4I​ = 3I4{{3I} \over 4}43I​

∴\therefore∴ New focal length = f

and intensity = 3I4{{3I} \over 4}43I​

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