Question

An equiconvex glass lens (i) has a focal length f and power P. It is cut into symmetrical halves. (ii) by a plane containing the principal axis. The two pieces are recombined as shown in figure (iii). The power of the new combination:

\[\begin{align}
  & \text{A) P} \\
 & \text{B) }\dfrac{\text{P}}{\text{2}} \\
 & \text{C) 2P} \\
 & \text{D) Zero} \\
\end{align}\]

Answer 1

Hint: We need to understand the relation of the focal length and the power of the lens with the shape of the lens. We can easily use the lens makers formula to find any possible relation between these quantities which can be used in this situation.

Complete answer:
We are given a special situation in which a given equiconvex lens is cut symmetrically along its axis as shown in the figure.

We need to check for the possible chances of change in the focal length or the power of the lens due to this. We know the lens makers formula which defines the focal length of a given lens. It is given as –
\[\dfrac{1}{f}=(\mu -1)(\dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}})\]
In our situation, the lens is a equiconvex lens,
\[\Rightarrow {{R}_{1}}=-{{R}_{2}}=R\]
We can rewrite the lens maker's formula as –
\[\dfrac{1}{f}=(\mu -1)(\dfrac{2}{R})\]
Now, let us consider cutting the given equiconvex lens into two symmetrical halves as shown in (ii). We know that there are now changes in the parameters involved in the lens maker's formula when cutting this along the axis. The radius of curvature of the lens remains the same. As a result, we can conclude that the power of the two new lenses formed from the initial one will be the same as that of it.
The power of the lens is given as –
\[P=\dfrac{1}{f}\]
So, the power of each lens in (ii) is also P.
Now, let us consider the new arrangement in which the two symmetrical pieces are placed as in (iii).
             

We can see that the first lens will try to converge with a power P, whereas the second lens will be diverging the same beam with the same power P. As a result of this, there will be no net power for the combination. The combination will act as plane glass.
The focal point of the combination of the two lenses is at infinity. The power of the new lens combination, is therefore, zero.

The correct answer is option D.

Note:
The power of a lens is the reciprocal of the focal length of the lens in meter scale. It is the degree of divergence or convergence of the lens. The higher the positive value of power, the greater will be the converging power of the given lens and vice versa.