Question

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away from a means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Answer 1

Hint: To manufacture the lens of suitable power and length the lens makers use the lens maker formula. Another formula used for the lens that relates the radii of curvature, focal length, and refractive index.

Formula used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ ; This is called the lens maker formula. Where f represents focal length, v is image distance and u are object distance.

Complete step by step solution:
Let object distance (u) be x and image distance (v) be y. so, according to the condition given in question. According to sign convention $u = - x$
We can write. $x + y = 3m$
From here we can find any one variable in terms of the other.
$y = 3 - x$ ………….. (1)
Let us now substitute the value in the formula $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
$\dfrac{1}{f} = \dfrac{1}{y} + \dfrac{1}{x}$
From equation (1) we can rewrite the equation as follows.
$\dfrac{1}{f} = \dfrac{1}{{3 - x}} + \dfrac{1}{x}$
 $ = \dfrac{{x + \left( {3 - x} \right)}}{{x\left( {3 - x} \right)}}$
Let us further simplify it.
$\dfrac{1}{f} = \dfrac{3}{{x\left( {3 - x} \right)}}$
$ \Rightarrow f = \dfrac{{x\left( {3 - x} \right)}}{3}$
Now, maximum possible focal length is asked. So, we will equate the first derivative of focal length f to zero to find its maximum value.
First, let us find the first derivative with respect to x.
$\dfrac{{df}}{{dx}} = \dfrac{{3 - 2x}}{3}$
Now let us equate it to zero.
$\dfrac{{df}}{{dx}} = \dfrac{{3 - 2x}}{3} = 0$
Let us further simplify it.
$3 - 2x = 0$
 $ \Rightarrow x = \dfrac{3}{2}m$
Hence, for maximum focal length object distance is 3/2. Now let us substitute the value of x in equation (1).
Hence, the value of image distance y is calculated as follows.
$y = 3 - \dfrac{3}{2} = \dfrac{3}{2}m$
That means both the distance of object and image is equal.
Now let us substitute the values of x and y in formula $\dfrac{1}{f} = \dfrac{1}{y} + \dfrac{1}{x}$and calculate the maximum focal length.
$\dfrac{1}{{{f_{\max }}}} = \dfrac{1}{{\dfrac{3}{2}}} + \dfrac{1}{{\dfrac{3}{2}}}$
$ = \dfrac{2}{3} + \dfrac{2}{3}$
On further simplifying we get the following.
\[\dfrac{1}{{{f_{\max }}}} = \dfrac{4}{3}\]
\[ \Rightarrow {f_{\max }} = \dfrac{3}{4}\]
\[ = 0.75m\]

$\therefore $ Hence, maximum focal length is 0.75m.

Note:
Another direct method; to obtain a real image on the wall minimum distance between images and objects should be equal to $4f$. Where $f$ is focal length. Here,
$4f = 3m$
$ \Rightarrow f = \dfrac{3}{4}m$