Question 393: Ray Optics

Question

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

Accepted Answer

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

Answer

$$f\,$$ and $${I \over 4}$$

Answer

$${{3f} \over 4}$$ and $${I \over 2}$$

Answer

$$f$$ and $${{3I} \over 4}$$

Answer

$${f \over 2}$$  and  $${I \over 2}$$

Ray Optics

A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter ${d \over 2}$ 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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