Question

A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. Aperture of diameter $\dfrac{d}{2}$​ in the central region of the lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively.
A) $f$ and $\dfrac{I}{4}$
B) $\dfrac{{3f}}{4}$ and $\dfrac{I}{2}$
C) $f$ and $\dfrac{{3I}}{4}$
D) $\dfrac{f}{4}$ and $\dfrac{I}{2}$

Answer 1

In this problem it is given that a lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. But as we know that the intensity $I$ of the image formed (or light) is directly proportional to the opening of the lens i.e., the area of aperture.
So, the wider the aperture, the more light will pass through and intensity of image will be more.

Step 1: As we know that the intensity of the image formed is directly proportional to the area of aperture of the lens receiving the light.
So, for the given condition when focal length is $f$ and diameter of aperture is $d$ then the area of aperture is given as follows-
$A = \pi \mathop r\nolimits^2$.................(1)
Where $r =$radius of aperture and here $r = \dfrac{d}{2}$
So, $A = \pi \mathop {\left( {\dfrac{d}{2}} \right)}\nolimits^2$..............(2)
And intensity $\propto$ Area
So, $I \propto A$ i.e. $I \propto \dfrac{{\mathop {\pi d}\nolimits^2 }}{4}$..................(3)

Step 2: But as it is given in the question that the aperture of diameter $\dfrac{d}{2}$ in the central region of the lens is covered by a black paper.
So, the available aperture is of diameter $\dfrac{d}{2}$ that will let pass the light.
The area of this new aperture is given by (let $\mathop A\nolimits_1$) –
$\mathop A\nolimits_1 = \pi \mathop {\left( {\dfrac{d}{2} \times \dfrac{1}{2}} \right)}\nolimits^2$..................(4)
And intensity for this new aperture (let $\mathop I\nolimits_1$) can be given as –
$\mathop I\nolimits_1 \propto \mathop A\nolimits_1$ i.e. $\mathop I\nolimits_1 \propto \dfrac{{\mathop {\pi d}\nolimits^2 }}{{16}}$.................(5)
Step 3: So, the available area that will let pass the light (let $\mathop A\nolimits'$) given by –
$\mathop A\nolimits' = A - \mathop A\nolimits_1$.................(6)
By using the values from equations (2) and (4) in (6), we will get –
$\mathop A\nolimits' = \pi \mathop {\left( {\dfrac{d}{2}} \right)}\nolimits^2 - \pi \mathop {\left( {\dfrac{d}{2} \times \dfrac{1}{2}} \right)}\nolimits^2$
$\mathop A\nolimits' = \pi \dfrac{{\mathop d\nolimits^2 }}{4}\left( {1 - \dfrac{1}{4}} \right) = \dfrac{{\mathop {3\pi d}\nolimits^2 }}{{16}}$
$\mathop I\nolimits' \propto \dfrac{{\mathop {3\pi d}\nolimits^2 }}{{16}}$...............(7)

Step 4: Now dividing the equation (7) by (3)
$\dfrac{{\mathop I\nolimits' }}{I} = \dfrac{{\mathop A\nolimits' }}{A} = \dfrac{{\mathop {3\pi d}\nolimits^2 }}{{16}} \times \dfrac{4}{{\mathop {\pi d}\nolimits^2 }}$
$\dfrac{{\mathop I\nolimits' }}{I} = \dfrac{3}{4}$
So, $\mathop I\nolimits' = \dfrac{3}{4}I$

Step 5: By converging the aperture, focal length cannot be changed on any lens. So focal length will remain the same as initial i.e., $f$.

So, the correct option is (C).

Note: The intensity of image formed by the lens also depends on the distance of the image from the lens.
With increase in the area of aperture here are a greater number of rays passing through the lens, so the intensity of the image will also increase and vice-versa.