Question

A thin lens of focal length f and its aperture has a diameter d. It forms an image of intensity I. Now the central part of the aperture upto diameter (d/2) is blocked by an opaque paper. The focal length and image intensity would change to.
(A) $\dfrac{f}{2},\dfrac{I}{2}$
(B) $f,\dfrac{I}{4}$
(C) $\dfrac{{3f}}{4},\dfrac{I}{2}$
(D) $f,\dfrac{{3I}}{4}$

Answer 1

Hint:In order to solve this problem firstly we calculate the area of aperture and area of blocked region . After them relate the area with intensity by following $[I\,\alpha \,A]$. Where, I = intensity of image and A= area of aperture.Now we can calculate the final intensity of image formed by the lens.For calculating the focal length change we will use following expression
$\dfrac{1}{f} = \left( {\mu - 1} \right)\,\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where $f$ = focal length of lens
$\mu = \dfrac{{{\mu _l}}}{{{\mu_s}}}\dfrac{{refractive\,index\,of\,lens\,material}}{{refractive\,index\,of\,surrounding\,medium}}$
${R_1}$= radius of cur water of surface 1
${R_2}$ = radius of cur water of surface 2

Complete step by step answer:
\[\dfrac{1}{f} = \left( {\mu - 1} \right)\,\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] …………1
In numerical given that the central part of the aperture is blocked but u,${R_1}$ and ${R_2}$ are not changed. So, according to equation 1, $f$ will not be changed hence the focal length of the lens will not be changed i.e. remains constant i.e. f .Now we have to find the change in intensity of the image. We know that the intensity is directly proportional to the area i.e.
$I\,\alpha \,A$
Initially given that the diameter of aperture is so the radius will be
$r = \dfrac{d}{2}$
So, the intensity of images is given as
\[I\,\alpha \,\pi \,{r^2}\]
$\Rightarrow r = \dfrac{d}{2}$
$\Rightarrow I\alpha \pi {\left( {\dfrac{d}{2}} \right)^2}$
$\Rightarrow I\alpha \dfrac{{\pi {d^2}}}{4}$ …………..2
Now, the control part of the aperture upto diameter $\left( {\dfrac{d}{2}} \right)$ is blocked. So, radius of this blocked portion is
${r^1} = \dfrac{{\dfrac{d}{2}}}{2}$
$\Rightarrow{r^1} = \dfrac{d}{4}$ …………..3
So the area of blocked portion is
${A^1} = \pi \,{r^1}^{}$
$\Rightarrow{A^1} = \pi {\left( {\dfrac{d}{4}} \right)^2}$
$\Rightarrow{A^1} = \dfrac{{\pi {d^2}}}{{16}}$
So, the area of remaining aperture from which image is formed is
${A_1} = A - {A^1}$
$\Rightarrow{A_1} = \dfrac{{\pi {d^2}}}{4} - \dfrac{{\pi {d^2}}}{{16}}$
$\Rightarrow{A_1} = \dfrac{{\pi {d^2}}}{4}\left( {1 - \dfrac{1}{4}} \right)$
$\Rightarrow{A_1} = \dfrac{{\pi {d^2}}}{4}\left( {\dfrac{{4 - 1}}{4}} \right)$
$\Rightarrow{A_1} = \dfrac{{3\pi {d^2}}}{{16}}$ …………..4
So, the intensity of image would change to
${I_1}\,\alpha \,{A_1}$
From eqn. 4
${I_1}\alpha \,\dfrac{{3\pi {d^2}}}{{16}}$ …………..5
On dividing eqn. 5 and 2
$\dfrac{{{I_1}}}{I} = \dfrac{{3\pi {d^2}}}{{16}} \times \dfrac{4}{{\pi {d^2}}}$
$\Rightarrow\dfrac{{{I_1}}}{I} = \dfrac{3}{4}$
So ${I_1} = \dfrac{3}{4}I$
Hence the changed intensity of image is $\dfrac{3}{4}I$ and focal length is f.

So option D is the correct answer.

Note: In many problem of intensity amplitudes is given so In that cases we have use following expression i.e,
$I\,\alpha \,{A^2}$ where I$ = $intensity
A $ = $ amplitude of light
And focal length of a lens does not depend on the size of aperture or blockage. It depends on the shape, size, nature and type of mirror and lens.