A short straight object of height 100 cm lies before the central axis of a spherical mirror, whose f

Q: Why can't it be a concave mirror?

Concave mirrors can indeed form virtual and erect images, but only when the object is placed very close to the mirror (between the pole and the focus). However, in that specific case, the image is always *magnified* ($m > 1$). Since our image is diminished ($m = 1/4$), it must be a convex mirror.

Q: What does 'same orientation' mean mathematically?

It means the image is erect (upright) relative to the object. Mathematically, this translates to a positive magnification ($m > 0$). For spherical mirrors, a positive magnification always corresponds to a virtual image.

Q: Why is the focal length of the convex mirror taken as positive?

By standard sign convention, the direction of incident light is taken as positive. For a convex mirror, the center of curvature and the principal focus lie behind the mirror, in the direction of the incident light. Therefore, their coordinates, and hence the focal length, are positive.

Q: Could the object be virtual?

The problem states the object 'lies before the central axis', which implies a standard real object placed in front of the mirror. Virtual objects are typically formed by converging rays from another optical element, which isn't the case here.

Q: What if the image had the opposite orientation?

If the image had the opposite orientation (inverted), the magnification would be negative, indicating a real image. Real images are formed on the same side as the object. In that case, the mirror would have to be concave, as convex mirrors cannot form real images of real objects.

Question

A short straight object of height 100 cm lies before the central axis of a spherical mirror, whose focal length has absolute value $f = 40	ext{ cm}$. The image of object produced by the mirror is of height 25 cm and has the same orientation of the object. One may conclude from the information.

Options:

(A) Image is real, same side of concave mirror

(B) Image is virtual, opposite side of concave mirror

(C) Image is real, same side of convex mirror

(D) Image is virtual, opposite side of convex mirror

Sigma Percentile · Accepted Answer

### The Clues in the Shadows

Imagine you are standing in front of a mysterious spherical mirror. You place a tall object, exactly $100	ext{ cm}$ in height, on the principal axis.

The mirror casts an image that is $25	ext{ cm}$ tall. But here is the most crucial piece of evidence: the image has the **same orientation** as the object.

In the realm of optics, an image that stands upright (same orientation) is always a **virtual image**. It is an optical illusion formed by the diverging rays of light, appearing to exist behind the mirror.

### The Magnification Revelation

Let's quantify this observation. The magnification $m$ of a mirror is the ratio of the image height $h_i$ to the object height $h_o$.

$$m = \frac{h_i}{h_o} = \frac{+25}{100} = +\frac{1}{4}$$

The positive sign mathematically confirms that the image is erect. More importantly, the value $\frac{1}{4}$ is less than $1$, which means the image is **diminished** (smaller than the object).

### The Identity of the Mirror

Now, we must play detective. Which type of spherical mirror can produce an image that is both virtual and diminished?

A **concave mirror** can indeed form a virtual image, but only when you stand very close to it. However, that virtual image is always magnified (like a makeup mirror).

A **convex mirror**, on the other hand, always scatters light outwards. No matter where you place a real object, a convex mirror will always form a **virtual, erect, and diminished** image. Therefore, our mysterious mirror is undeniably convex!

### The Mathematical Proof

Let's solidify our logical deduction with rigorous mathematics. We know the absolute focal length is $40	ext{ cm}$. For a convex mirror, the focus lies behind the mirror, so $f = +40	ext{ cm}$.

Using the magnification formula in terms of focal length:

$$m = \frac{f}{f - u}$$

Substituting our known values:

$$\frac{1}{4} = \frac{40}{40 - u}$$

Solving for the object distance $u$:

$$40 - u = 160 \implies u = -120	ext{ cm}$$

Now, let's find the exact location of the image using the standard magnification relation:

$$m = -\frac{v}{u}$$

$$\frac{1}{4} = -\frac{v}{-120}$$

$$v = +30	ext{ cm}$$

The positive sign of $v$ is the final nail in the coffin. It proves that the image is formed $30	ext{ cm}$ **behind the mirror**, on the opposite side of the object.

Thus, we can confidently conclude that the image is virtual and located on the opposite side of a convex mirror.

Select Answer:

Visualized Solution

$h_{o} = 100 cm$

$m = \frac{h _{i}}{h _{o}}$

\text{Mirror Type}

\text{Ray Diagram}

\text{Mathematical Verification}

\text{Image Position}

The Sigma Insight: Spherical Mirror

The Clues in the Shadows

The Magnification Revelation

The Identity of the Mirror

The Mathematical Proof

Similar Questions

A spherical mirror is obtained as shown in the above figure from a hollow glass sphere. If an object is in front of the mirror, what will be the nature and magnification of the image of the object? (Figure drawn as schematic and not to scale)

A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let $O$ be the pole of the mirror and $C$ its centre of curvature. A point object is placed at $C$ . It has a real image, also located at $C$ . If the mirror is now filled with water, the image will be

A short linear object of length $b$ lies along the axis of a concave mirror of focal length $f$ at a distance $u$ from the pole of the mirror. The size of the image is approximately equal to

A container is filled with water ( $μ = 1.33$ ) up to a height of $33.25 cm$ . A concave mirror is placed $15 cm$ above the water level and the image of an object placed at the bottom is formed $25 cm$ below the water level. The focal length of the mirror is

A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length $f$ as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale.)

A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire ? (These figures are not to scale.) ?

An object is placed beyond the centre of curvature $C$ of the given concave mirror. If the distance of the object is $d_{1}$ from $C$ and the distance of the image formed is $d_{2}$ from $C$ , the radius of curvature of this mirror is

A thin rod of length $\frac{f}{3}$ is placed along the optic axis of a concave mirror of focal length $f$ such that its image which is real and elongated, just touches the rod. The magnification is ......

When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of $9 cms^{- 1}$ , the speed (in $cms^{- 1}$ ) with which image moves at that instant is ........... .