Question

Use the mirror formula to prove that:
(a) An object placed between f and 2f of a concave mirror produces a real image beyond 2f
(b) A convex mirror always produces a virtual image independent of the location of the object.
(c) The virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Answer 1

Hint: This question is the easy application of the mirror formula. You have to use right sign convention and put the values (that are given to you in the question) in the mirror formula. After that, find the relation between u, v, and f to answer each part. Also solve the question part by part.

Complete answer:
(a) The focal length (f) is negative for a concave mirror. If the object is located on the left side of the mirror, the distance from the object (u) is negative. We should write for Image Distance v
Then
By the mirror formula
We have,
\[\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\]
So,
\[\Rightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\] …………… (1)
The object is placed between f and 2f.
2f < u < f
So,
Relation will be opposite for the reciprocals
$\dfrac{1}{2f}>\dfrac{1}{u}>\dfrac{1}{f}$
Multiplying each by -1
We have
$-\dfrac{1}{2f}<-\dfrac{1}{u}<-\dfrac{1}{f}$
So,
$\dfrac{1}{f}-\dfrac{1}{2f}<\dfrac{1}{f}-\dfrac{1}{u}$ …………………… (2)
Using the eq. 1
We have
$\dfrac{1}{2f}<\dfrac{1}{v}<0$; v is negative
$2f>v$
So,
$-v>-2f$
Therefore, the image lies beyond 2f.

(b) The focal length (f) is positive for a convex mirror. If the object is located on the left side of the mirror, the distance from the object (u) is negative. We have the mirror formula for reflection distance v
Now,
\[\begin{align}
  & \dfrac{1}{\Rightarrow v}+\dfrac{1}{u}=\dfrac{1}{f} \\
 & \Rightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u} \\
\end{align}\]
From eq. 2
We have
$\dfrac{1}{v}<0$
Therefore,
v > 0
On the back side of the mirror, the picture is thus created. Therefore, a convex mirror, regardless of the object distance, still creates a virtual image.

(c) The focal length (f) is positive for a convex lens.
If the object is located on the left side of the mirror, the distance from the object (u) is negative.
We have the mirror formula for reflection distance v
\[\begin{align}
  & \dfrac{1}{\Rightarrow v}+\dfrac{1}{u}=\dfrac{1}{f} \\
 & \Rightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u} \\
\end{align}\]
But,
In this case
u < 0
so,
$\dfrac{1}{v}>\dfrac{1}{f}$
And,
f > v
The picture created is then reduced and is positioned between the focus (f) and the pole.

(d) The focal length (f) for a concave mirror is negative. If the object is located on the left side of the mirror, the distance from the object (u) is negative. Between the emphasis (f) and the pole it is positioned
Hence, f > u > 0
$\dfrac{1}{f}<\dfrac{1}{u}<0$
Therefore
$\dfrac{1}{f}-\dfrac{1}{u}<0$
For image distance v, we have the mirror formula
\[\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\]
So,
\[\Rightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\]
So,
$\dfrac{1}{v}<0$
v > 0
On the right hand side of the mirror, the picture is created. Therefore, it is a virtual picture that
We should write: For u <0 and v > 0:
$\dfrac{1}{u}>\dfrac{1}{v}$
So,
v > u
therefore
magnification = $\dfrac{v}{u}>1$
Hence, the formed image is enlarged.

Note:
Sign convention is a set of guidelines for mathematical study of image creation to set signs for image distance, object distance, focal length, etc. It's according to:
1. Objects are typically located to the left of the mirror
2. From the pole of the mirror, all lengths are determined.
3. The distances measured are positive in the direction of the incident ray, and the distances measured are negative in the direction opposite to that of the incident ray.
4. Distances measured above the main axis along the y-axis are positive, and those measured below the main axis along the y-axis are negative.
5. It is possible to reverse the sign convention and it would always provide the right results.