A lens having focal length \(f\) and aperture of diameter \(d\) forms an image of intensity \(I\). An aperture of diameter \(\frac{d}{2}\) in central region of lens is covered by a black paper. The focal length of lens and intensity of the image now will be respectively:
1. \(f\) and \(\frac{I}{4}\)
2.  \(\frac{3f}{4}\) and \(\frac{I}{2}\)
3.  \(f\) and \(\frac{3I}{4}\)
4.  \(\frac{f}{2}\) and \(\frac{I}{2}\)

An object is placed at a point distance \(x\) from the focus of a convex lens and its image is formed at \(I\) as shown in the figure. The distances \(x\) and \(x'\) satisfy the relation:

          
1. \(\frac{x+x'}{2} = f\)
2. \(f = xx'\)
3. \(x+x' \le 2f\)
4. \(x+x' \ge 2f\)

A thin equiconvex lens of power \(P\) is cut into three parts \(A,B,\) and \(C\) as shown in the figure. If \(P_1,P_2\) and \(P_3\) are powers of the three parts respectively, then:
            

A concave lens forms the image of an object such that the distance between the object and image is \(10\) cm. If magnification of the image is \(\frac{1}{4},\) the focal length of the lens is:
1. \(-\frac{20}{3}~\text{cm}\)
2. \(\frac{20}{3}~\text{cm}\)
3. \(\frac{40}{9}~\text{cm}\)
4. \(-\frac{40}{9}~\text{cm}\)

In the diagram shown below, the image of the point object \(O\) is formed at \(l\) by the convex lens of focal length \(20~\text{cm},\)  where \(F_1\) and \(F_2\) are foci of the lens. The value of \(x'\) is: