Question

A double convex lens of focal length \[f\] is cut into \[4\] equivalent parts. One cut is perpendicular to the axis and the other is parallel to the axis of the lens. The focal length of each part is:
A. \[f/2\]
B. \[f\]
C. \[2f\]
D. \[4f\]

Answer 1

Hint: First of all, we will find the focal length of the double convex lens using lens maker formula assuming both the radius of curvature to be the same. After that we find the focus of the cut lens using the same formula, but different radius of curvature as per requirement. We will substitute the required values and manipulate.

Complete step by step answer:In the given problem, we are supplied the following data:
The focal length of a double convex lens is \[f\] .
It is cut into four equivalent parts, such that one cut is perpendicular to the axis and the other cut is parallel to the axis.
We are asked to find the focal length of each part.

For this we will apply the principle that when the lens is cut parallel to the axis, there is no change in the focal length of the respective lens. The radius of curvature will remain the same when it is cut parallel to the axis, and hence the focal length also remains the same. In contrast to this, when the lens is cut perpendicular to the axis the focal length changes which will be different from the previous value. In case the lens is cut, the intensity of the image will be less than that of previous intensity.

Now, we apply lens maker formula,
\[\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] …… (1)
Where,
\[f\] indicates the focal length of the lens.
\[\mu \] indicates a refractive index.
\[{R_1}\] indicates the radius of curvature of one face.
\[{R_2}\] indicates the other.
When we cut the lens parallel to the axis, there is no change in the radius of curvature.
So, we can take the magnitude of the radius of curvature to be same, as follows:
\[{R_1} = {R_2} = R\]
However,
\[{R_2} = - R\] , due to orientation of the face of the lens (sign convention).
Now, we substitute the required values in equation (1), we get:
\[
\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{R} - \left( {\dfrac{1}{{ - R}}} \right)} \right) \\
\Rightarrow \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{R} + \dfrac{1}{R}} \right) \\
\Rightarrow \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{2}{R}} \right) \\
 \Rightarrow f = \dfrac{R}{{2\left( {\mu - 1} \right)}} \\
\]
But when the lens is cut parallel to the axis, then one of the face’s radius of curvature becomes infinity, because of the planar face.
So, we can write:
\[
  {R_2} = \, \propto \\
\Rightarrow \dfrac{1}{{{R_2}}} = \dfrac{1}{ \propto } \\
\Rightarrow \dfrac{1}{{{R_2}}} = 0 \\
\]
Again, we use the lens maker formula:
\[
\dfrac{1}{{{f_1}}} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\
\Rightarrow \dfrac{1}{{{f_1}}} = \left( {\mu - 1} \right)\left( {\dfrac{1}{R} - 0} \right) \\
\Rightarrow \dfrac{1}{{{f_1}}} = \left( {\mu - 1} \right) \times \dfrac{1}{R} \\
\Rightarrow {f_1} = \dfrac{R}{{\left( {\mu - 1} \right)}} \\
\]
Again, manipulating the above expression, we get:
\[
  {f_1} = \dfrac{R}{{\left( {\mu - 1} \right)}} \\
\Rightarrow {f_1} = 2 \times \dfrac{R}{{2\left( {\mu - 1} \right)}} \\
\Rightarrow {f_1} = 2f \\
\]
Hence, the focal length of each part is found to be \[2f\] .

The correct option is C.

Note:While solving this problem, we should keep in mind that in case of double convex lens, the radius of curvature will be same, but when the lens is cut perpendicular to the axis, one of the faces becomes planar, whose radius of curvature is present at infinity. This is the same as that of the radius of curvature of a plane mirror, whose radius of curvature is also present at infinity due to planar face. This should not be confused as it is the main part of the solution where you may go wrong.