Question

The minimum distance between an object and its real image formed by a convex lens is
(A) $\dfrac{2}{3}f$
(B) $2f$
(C) $\dfrac{5}{2}f$
(D) $4f$

Answer 1

Hint: We use the lens equation to solve this problem. We first write the distance from image to lens in terms of distance of object from the lens and distance between object and real image. We substitute this in the lens equation to find the distance between an object and its real image.

Formula used: lens equation of convex lens
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Here,
Focal length of lens is$f$
Distance from image to lens is represented by $v$
Distance between lens and object is represented by $u$

Complete step by step solution:
Let the distance between object and real image be$d$.
Let the distance of image from the lens be $y$
So, the distance from object to lens is $d - y$
Substituting this in the lens equation,
\[u = - (d - y),v = + y\]
\[\dfrac{1}{f} = \dfrac{1}{y} - ( - \dfrac{1}{{d - y}})\]
Solving for $d$
\[\dfrac{1}{f} = \dfrac{1}{y} + \dfrac{1}{{d - y}}\]
${y^2} - yd - yf = 0$
\[y = \dfrac{{d \pm \sqrt {{d^2} - 4df} }}{2}\]
For $y$ to be real
${d^2} > 4fd$
$d > 4f$
So minimum distance between object and image should be $d > 4f$

Hence option (D) $d > 4f$ is the correct answer

Additional information: A convex lens is also known as a converging lens. A converging lens is a lens that converges rays of light that are traveling parallel to its principal axis. They can be identified by their shape, they are thick at the centre and thin at the edges. They are also called positive lenses.

Note: We can also solve this question by drawing a ray diagram of convex lens when the object is located at $2f$then the image formed will be at $2f$ on the other side of the lens. Adding the two distances we get$4f$ as the answer.
We take \[u = - (d - y)\]as negative because of sign conventions, as we are going from right to left.