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 –

Answer 1

Answer:

Explanation:

The focal length of the lense does not changes even after covering with black paper. Thus, the focus will be f.

We know that the intensity is directly proportional to the square of area and this we can write as intensity (I) directly proportional to area(A)^2.

As area of the lense we can write by (pie*r^2) then for aperture of diameter d/2 we get the area to be (pie*(r/2)^2).

Since, the area is (pie*(r/2)^2).

Thus, for area2 (pie*(r/2)^2)/4 we get that the area2 will be (pie*(r)^2)/16.

Hence, the net area will be area1 - area2.

(pie*(r)^2)/4 - (pie*(r)^2)16

=3(pie*(r)^2)/16

Thus, now to find the intensity I1/I2 will be A1/A2.

= [(pie*r^2)/4]/3(pie*(r)^2)/16

=3/4.

Hence, I2 = 4/3(I1) and focus will remain f.