Question

A thin lens forms an image of intensity $I$. If the central part of the lens diameter half the diameter of the lens is black polished, then what will be the intensity of the image formed by the remaining part?

Answer 1

Use the formula for area of the circle in terms of diameter of the circle. Using this formula first determine the initial area of the lens and black polished area of the lens. Then calculate the area of the lens which is unpolished. Since the intensity of the image formed is directly proportional to the area of the lens forming the image, determine the intensity of the image formed by the unpolished area.

Formula used:
The area $A$ of a circle is given by
$A = \dfrac{{\pi {d^2}}}{4}$ …… (1)
Here, $d$ is the diameter of the circle.

Complete step by step answer:
We have been given a thin lens that forms an image with intensity $I$. Let $d$ be the initial diameter of the lens. According to equation (1), the initial area of the lens is
$A = \dfrac{{\pi {d^2}}}{4}$
We have given that the central part of the lens having a diameter equal to half of the initial diameter is black polished. Hence, the diameter of the black polished circle is
$d' = \dfrac{d}{2}$

Let us first determine the area of the black polished lens. According to equation (1), the black polished area of the lens is
$A' = \dfrac{{\pi {d^{'2}}}}{4}$
Substitute $\dfrac{d}{2}$ for $d'$ in the above equation.
$A' = \dfrac{{\pi {{\left( {\dfrac{d}{2}} \right)}^2}}}{4}$
$\Rightarrow A' = \dfrac{{\pi {d^2}}}{{16}}$

The area $A''$ of the lens which is unpolished is
$A'' = A - A'$
$\Rightarrow A'' = \dfrac{{\pi {d^2}}}{4} - \dfrac{{\pi {d^2}}}{{16}}$
$\Rightarrow A'' = \dfrac{{4\pi {d^2} - \pi {d^2}}}{{16}}$
$\Rightarrow A'' = \dfrac{{3\pi {d^2}}}{{16}}$
This is the unpolished area of the lens.

The intensity of the image formed by the lens initially is $I$. We have asked to determine the intensity $I'$ of the image formed by the unpolished area of the lens. We know that the intensity of the image formed is directly proportional to area of the lens contributing in the image formation.
$I \propto A$
We can write this relation for the two intensities of the image before and after polishing the lens as
$\dfrac{{I'}}{I} = \dfrac{{A''}}{A}$

Substitute $\dfrac{{\pi {d^2}}}{4}$ for $A$ and $\dfrac{{\pi {d^2}}}{{16}}$ for $A''$ in the above equation.
$\dfrac{{I'}}{I} = \dfrac{{\dfrac{{3\pi {d^2}}}{{16}}}}{{\dfrac{{\pi {d^2}}}{4}}}$
$\Rightarrow \dfrac{{I'}}{I} = \dfrac{{12}}{{16}}$
$\therefore I' = \dfrac{{3I}}{4}$

Hence, the intensity of the image formed by the remaining part of the lens is $\dfrac{{3I}}{4}$.

Note: The students may forget to determine the area of the unpolished surface of the lens and determine the intensity of the image formed by the polished surface in terms of initial intensity of the image. But the students should keep in mind that there is no image formation by the black polished area of the lens. So, the answer will be incorrect. Also the students may think that the image will not be formed if some area of the lens is polished. But the image is formed with less intensity.