Animated Solution for Physics - Optics: Which of the following form(s) a virtual and erect image for all positions of the object ?

In the fascinating world of optics, predicting the nature of an image formed by lenses and mirrors is a fundamental skill. This question challenges us to identify which optical instruments always produce a virtual and erect image for any position of a real object. Let's embark on a mathematical and visual journey to uncover the truth!

Let's start by examining the concave lens. The behavior of any lens is governed by the elegant lens formula:

To understand where the image forms, we can rearrange this equation to solve for the image distance, $v$:

Now, we must apply the Cartesian sign convention. For a concave lens, the principal focus lies on the same side as the incident light, which means its focal length $f$ is inherently negative ($f<0$). Furthermore, for any real object placed in front of the lens, the object distance $u$ is also negative ($u<0$).

Let's substitute these negative values into our rearranged formula:

Look closely at this result! The term inside the parentheses is the sum of two positive numbers, making it strictly positive. Because of the negative sign outside, $\frac{1}{v}$ is always negative. Consequently, the image distance $v$ is always negative.

In the realm of lenses, a negative image distance signifies that the image is formed on the same side as the object. Such an image is, by definition, virtual and erect. Thus, a concave lens will always yield a virtual image for a real object.

Now, let's shift our focus to the convex mirror. The governing equation here is the mirror formula:

Again, we rearrange this to isolate the image distance, $v$:

Applying the sign convention for a convex mirror reveals a different setup. A convex mirror curves outwards, placing its principal focus behind the reflective surface. Therefore, its focal length $f$ is positive ($f>0$). As always, for a real object, the object distance $u$ remains negative ($u<0$).

Substituting these signs into our equation gives:

Here, the two negative signs cancel out, leaving us with the sum of two positive quantities. This guarantees that $\frac{1}{v}$ is always positive, which in turn means that $v$ is always positive.

For mirrors, a positive image distance indicates that the image is formed behind the mirror. An image formed behind a mirror is always virtual and erect. Hence, a convex mirror, much like a concave lens, will always produce a virtual image for a real object.

Through rigorous mathematical analysis using sign conventions, we have proven that both the concave lens and the convex mirror are steadfast in their behavior: they will always form a virtual, erect, and diminished image for any real object, regardless of its position. Therefore, the correct options are (b) and (c).