Question

For a thin convex lens, when the height of the object is doubled then its image, its object distance is equal to
$\begin{align}
  & A)f \\
 & B)2f \\
 & C)3f \\
 & D)4f \\
\end{align}$

Answer 1

Hint: Magnification of a thin convex lens is given by the ratio of image distance to the object distance. It is also equal to the ratio of the size of image to the size of object. Lens formula says that reciprocal of focal length of the lens is equal to the difference in reciprocal of object distance from the reciprocal of image distance. From all these facts, object distance of the given thin convex lens can be determined.

Formula used: $1)M=\dfrac{{{h}_{i}}}{{{h}_{o}}}=-\dfrac{v}{u}$
$2)\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Complete step by step answer:
We know that magnification of a thin convex lens is nothing but a measure of how large, the object has been enlarged or magnified by the lens. Magnification of a lens can be defined as the ratio of size or height of the image of an object to the size or height of the object itself. Mathematically, it is given by
$M=\dfrac{{{h}_{i}}}{{{h}_{o}}}$
where
$M$ is the magnification of a lens
${{h}_{i}}$ is the height or size of the image of an object when viewed through the lens
${{h}_{o}}$ is the height or size of the object itself
Let this be equation 1.
We also know that magnification of a lens is given by the ratio of image distance from the lens to the object distance from the lens. Mathematically, this is given by
$M=-\dfrac{v}{u}$
where
$M$ is the magnification of a lens
$v$ is the image distance from the lens
$u$ is the object distance from the lens
Let this be equation 2.
Combining equation 1 and equation 2, we have
$M=\dfrac{{{h}_{i}}}{{{h}_{o}}}=-\dfrac{v}{u}\Rightarrow v=-\dfrac{u{{h}_{i}}}{{{h}_{o}}}$
Let this be equation 3.
Now, we know that lens formula is given by
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
where
$f$ is the focal length of a lens
$v$ is the image distance from the lens
$u$ is the object distance from the lens
Let this be equation 4.
Coming to our question, we are given a thin convex lens, wherein height of the object is doubled then its image. We are required to determine the object distance from the given lens.
If we call height of the object ${{h}_{o}}$, height of the image ${{h}_{i}}$, image distance $v$ and object distance $u$, then, using equation 3, image distance of the given thin convex lens is given by
$v=-\dfrac{u{{h}_{i}}}{{{h}_{o}}}\Rightarrow v=-\dfrac{u}{2}$
where
$u$ is the object distance from the given convex lens
$v$ is the image distance from the given convex lens
$\dfrac{h{}_{i}}{{{h}_{o}}}=\dfrac{1}{2}$, is the condition, as provided in the question
Let this be equation 5.
Now, substituting equation 5 in equation 4, we have
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\Rightarrow \dfrac{1}{f}=\dfrac{1}{\left( \dfrac{-u}{2} \right)}-\dfrac{1}{u}\Rightarrow \dfrac{1}{f}=-\dfrac{2}{u}-\dfrac{1}{u}=\dfrac{-3}{u}\Rightarrow u=-3f$
Distance can’t be negative so we will take magnitude of this.

So, the correct answer is “Option C”.

Note: Lens formula as well as the formula for magnification follows sign convention. Object distance is usually taken as negative whereas focal length of the lens is taken as positive, for easiness. It is due to this reason that equation 2, equation 3 and the final answer are having negative signs in their corresponding expressions. Thus, there is no need to worry about the negative sign in the final answer and students can confidently proceed by choosing option $C$, which gives the magnitude of object distance from the given thin convex lens.