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(D) The intensity of the image formed by a lens is proportional to the power collected by the lens and inversely proportional to the area of the image

Vedclass · Accepted Answer

$4I$

Vedclass · Answer

(D) The intensity of the image formed by a lens is proportional to the power collected by the lens and inversely proportional to the area of the image. The power collected by the lens is proportional to the area of the lens aperture,which remains constant since the diameter is the same. The area of the image $A$ is proportional to the square of the focal length $(A \propto f^2)$. Therefore,the intensity $E$ is proportional to $1/f^2$. Initially,$I \propto 1/f_1^2 = 1/(30)^2$. Finally,$I' \propto 1/f_2^2 = 1/(15)^2$. Taking the ratio: $I'/I = (f_1/f_2)^2 = (30/15)^2 = 2^2 = 4$. Thus,the new photoelectric current is $I' = 4I$.

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