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 \(\frac{I}{2}\)

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:
            

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:
        

If the space between two convex lenses of glass in the combination shown in the figure below is filled with water, then:
               

The distance between a convex lens and a plane mirror is \(10\) cm. The parallel rays incident on the convex lens, after reflection from the mirror form image at the optical centre of the lens. Focal length of the lens will be:

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. \(x+x' \ge 2f\)