Q.
A biconvexlens is cut intotwo equal parts by a plane perpendicular to the principal axis. If the power of the original lens is 4D, the power of a cut lens will be
Focal length o f a biconvex lens, f1=(μ−1)(R11−R21)
Here, R1=+R,R2=−R ∴f1=(μ−1)(R1−(−R)1) f1=R2(μ−1)
or f=2(μ−1)R…(i)
When a biconvex lens is cut into two equal parts by a plane perpendicular to the
principal axis, then each half behaves as a piano convex lens.
Focal length o f a piano convex lens f′1=(μ−1)(R11−R21)
Here R1=+R,R2=∞ ∴f′1=(μ−1)(R1)
or f′=(μ−1)R…(iii)
From (i) and (ii), we get f′=2f…(iii)
Power of a lens p=f1…(iv) ∴p′=f′1=2f1(Using(iii)) =2p p′=2p=24D=2D (GivenP=4D)