Question

A concave mirror has a focal length of \[20cm\]. The distance between the two positions of the object for which the image size is double of the object size is:
\[\begin{align}
  & A.20cm \\
 & B.40cm \\
 & C.30cm \\
 & D.60cm \\
\end{align}\]

Answer 1

Hint: The mirror formula is the relationship between the distance of an object $u$, distance of image $v$, and the focal length of the lens $f$. This law can be used for both concave and convex mirrors with appropriate sign conventions.
Formula Used:
Mirror formula: $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
Where, $f$ is the focal length of the mirror,$u$ is the object distance and $v$ is the image.

Complete step-by-step solution:
Here given that the focal length of the concave mirror is \[20cm\], i.e. $f=-20cm$
We know that the concave mirror can produce either real or virtual images, clearly there are two positions at which the size of image is twice the size of the object.
To begin with, let us find the image distance.
Since , it is given that the size of the image is twice the size of the object. Then, we can say that, $m=\dfrac{h_{i}}{h_{o}}=\dfrac{-v}{u}$
Then, $\dfrac{2h_{o}}{h_{o}}=\dfrac{-v}{u}$
Or$\dfrac{-v}{u}=2$
Or $v=-2u$
Let us consider the case where the real image is formed.
Then we know from mirror formula that, $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
Then, we have $u=-u$ and $v=-2u$
Substituting the values we get, $\dfrac{1}{-2u}-\dfrac{1}{u}=\dfrac{1}{-20}$
Then reducing we get, $\dfrac{3}{-2u}=\dfrac{1}{-20}$
Then we have, $u=30cm$
Similarly, Let us consider the case where the virtual image is formed.
Then we have $u=-u$ and $v=2u$
Substituting the values we get, $\dfrac{1}{2u}-\dfrac{1}{u}=\dfrac{1}{-20}$
Then we have, $\dfrac{-1}{2u}=\dfrac{1}{-20}$
Then, $u=10cm$
Now we have $u=30cm$ and $u=10cm$, then the difference between the two is \[20cm\]
Hence, the answer is A.\[20cm\]

Note: To identify the nature of the object, like magnification, magnification equation is used which states $M=\dfrac{Height\; of \;image}{Height\; of \;object}=-\dfrac{distance\; of\; image}{distance\; of\; object}$ if $M=+$ then the image is magnified and if $M=-$ then the image is diminished. Here we have two cases, where the image produced is either a real or virtual image depending on where the object is placed.