Q.A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. 9.7. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror?

Unlike the usual object standing perpendicular to the axis (all points at one $u$), a phone lying along the axis has its near end $B$ and far end $E$ at different object distances $u_B \ne u_E$.

1. Locate each end's image separately with the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$; since $u_B\ne u_E$, the resulting image distances (and magnifications $m=-v/u$) differ for the two ends.
2. Different magnification along the length means the image is stretched/compressed unevenly - not a faithful scaled copy of the phone.
3. Location matters: far from the mirror, $u_B\approx u_E$ so distortion is small; close to the mirror - especially with one end inside $F$ and the other outside - the distortion becomes severe (one image virtual, one real).

Setting up the diagram

Draw a concave mirror with pole $P$, principal axis, focus $F$, and centre of curvature $C$ marked on the axis. Unlike the usual textbook object (a small arrow standing perpendicular to the axis, where every point is at the same distance from the mirror), here the "object" - the phone - lies along the axis itself. Label its near end $B$ (closer to the mirror, at object distance $u_B$) and its far end $E$ (farther away, at object distance $u_E$), with $u_B < u_E$ in magnitude.

Locating the image of each end

Use the mirror formula $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ (magnitudes, concave mirror) separately for each end, since each is effectively a point object on the axis:

• Near end $B$: solve for $v_B$ using $u=u_B$. If $B$ lies between the pole and the focus ($u_B<f$), $v_B$ comes out negative - a virtual image, behind the mirror.
• Far end $E$: solve for $v_E$ using $u=u_E$. If $E$ lies beyond the focus, $v_E$ is positive - a real image, in front of the mirror.

Draw $B'$ and $E'$ at these two computed image positions on the axis; the image of the phone is the segment $B'E'$.

Why the magnification is not uniform

For a point on the axis, the lateral magnification is $m=-v/u$. Because $u$ is different for $B$ and $E$, the resulting $v$ (and hence $m$) is different for each - so the image length $B'E'$ is not simply a uniformly-scaled copy of the real length $BE$. (Any dimension of the phone that lies genuinely perpendicular to the axis, at a single value of $u$, would still image with one single magnification - it's specifically the along-the-axis extent that gets distorted, because that's the direction along which $u$ itself varies.)

Since $u_B\ne u_E$, in general $m_B \ne m_E$ - this unevenness in magnification along the length of the phone is exactly what produces the distorted image (nose-and-ears-style stretching, the same effect that distorts a selfie taken at close range with a tilted phone).

Does the distortion depend on where the phone is placed?

Yes.

So moving the phone closer to the mirror (especially straddling the focus) makes the distortion worse; moving it far away makes the distortion shrink toward zero.

Method: Imaging an Object Extended Along the Principal Axis

This method applies whenever an object does not stand perpendicular to the axis at a single distance, but instead extends along the axis itself (a rod, a phone, a pencil lying flat toward the mirror).

Steps

Step 1: Notice that the usual magnification formula assumes one fixed $u$

The mirror equation $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ and the magnification $m=-v/u$ are written for a single object distance $u$. The moment an object has different points at different distances from the pole, you can no longer treat it as one point object.

Step 2: Break the extended object into representative points

Pick the two (or more) extreme points of the object — e.g. the near end and the far end — and treat each as its own point object with its own object distance ($u_{\text{near}}$, $u_{\text{far}}$, ...).

Step 3: Apply the mirror equation separately to each point

Solve for the image distance $v$ of each point independently, using the sign convention consistently. Note whether each resulting image is real or virtual (compare $u$ to $f$).

Step 4: Compare the magnifications, not just the image positions

Compute $m=-v/u$ for each point separately. Because $u$ differs across the object, $m$ differs too — this unevenness in magnification (not the image positions alone) is what produces a stretched/distorted image rather than a faithful scaled copy.

Step 5 (Applying to this problem): Reason about how the distortion changes with location

Check how close the object's points sit to the focus. Points straddling $F$ give one real and one virtual image (severe, discontinuous-looking distortion); points far from the mirror give nearly equal $u$ values for both ends (small distortion). This qualitative check — without recomputing exact numbers — answers "does the distortion depend on location" questions directly.