Question

A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing
a) a concave mirror of suitable focal length
b) a convex mirror of suitable focal length
c) a convex lens of focal length less than 0.25m
d) a concave lens of suitable focal length

Answer 1

Hint: The first step should be to check which lens or mirror among the ones given in the options can produce a real and diminished image of the object. Then we can form an equation using the condition given that the distance between object and its image is 1.0 m. Then by solving that equation we will get the required answer.

Complete step by step solution:
Let the distance of the object from the mirror or lens be $u$ and image distance be $v$ and the focal length be $f$.
A real and diminished image of an object can be formed on a screen by the convergence of light rays coming from an object and is possible only with a convex lens or a concave mirror and placing the object at a distance beyond twice the focal length of the lens or mirror.
Suppose the object is at a distance of $x$ units beyond $2f$.
Thus, we can write that $u=x+2f$ ………. (i)
From the lens formula, we have $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\implies \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{2f+x}$
$\implies \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{2f+x}$
Upon solving and rearranging the terms, we will get
$v=\dfrac{f(2f+x)}{f+x}$ ………. (ii)
We have been given that the image of the object is at 1.0m distance from the object. Which means $u+v=1$
Now, inserting the value of $u$ and $v$, we get
$2f+x+\dfrac{f(2f+x)}{f+x}=1$
From, the above equation, we can write the inequality
$\dfrac{(2f+x)^2}{f+x}<1$
$\implies (2f+x)^2 <{f+x}$
This is possible only when the focal length, $f < 0.25$m.
Hence, option c is the correct answer.

Note: The characteristics of the image formed will decide which mirror or lens we can use, so knowledge about different lenses and mirrors is must. Another approach to solve this question can be by using the fact that for a convex lens the minimum distance between the object and its real image is four times the focal length.