A lens having focal length f and aperture of diameter d forms an image of intensity I . Aperture of diameter $$d2$$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

Question

A lens having focal length f and aperture of diameter d forms an image of intensity I . Aperture of diameter $$d2$$  in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

Infinity Learn · Accepted Answer

Focal length of the lens remains same. Intensity of image formed by lens is proportional to area exposed to incident light from object. ∴$$I2I1=A2A1Initially$$,   $$A1$$ $$=$$ π$$d22$$  $$=$$ π$$d24Finally$$, after blocking exposed area is $$A2$$ $$=$$ π$$d24$$ − π$$(d4)2$$ $$=$$ $$3$$$$π$$$$d216$$∴$$I2I1=A2A1$$ $$=$$  $$3$$$$π$$$$d216$$$$π$$$$d24$$ $$=$$ $$34$$ ⇒$$I2$$ $$=$$  $$34I1$$ $$=$$ $$34IHence$$, focal length of a lens $$=$$ f    and     Intensity of the image $$=$$ $$3I/4$$

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:

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A lens having focal length f and aperture of diameter d forms an image of intensity I . Aperture of diameter d2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

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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 respectively: