Mirror Equation of Spherical Mirrors

An image formed when reflected rays actually converge at a point. A real image can be obtained on a screen and is located in front of the mirror.

An image formed when reflected rays do not actually meet, but appear to diverge from a point. A virtual image cannot be obtained on a screen and is located behind the mirror.

To locate an image in a spherical mirror, any two of the following standard rays are used:

1. A ray parallel to the principal axis — after reflection- passes through the principal focus F (concave) or appears to diverge from F (convex).
2. Ray through the principal focus — after reflection- travels parallel to the principal axis.
3. Ray through the centre of curvature C — Retraces its path after reflection (falls normally on the mirror surface).
4. Ray incident at the pole is reflected according to the laws of reflection and makes equal angles with the principal axis.

All distances are measured from the pole P of the mirror.

Key results from sign convention:

• Object distance u is always negative for a real object placed in front of the mirror.
• Focal length f is negative for a concave mirror (focus is in front).
• Focal length f is positive for a convex mirror (focus is behind the mirror).
• Image distance v is negative if the image is real (in front), positive if the image is virtual (behind the mirror).

\[\frac {i}{v}\] + \[\frac {1}{u}\] = \[\frac {1}{f}\]

where:

• v = image distance (measured from the pole)
• u = object distance (measured from the pole)
• f = focal length of the mirror

Relation between focal length and radius of curvature: f = \[\frac {R}{2}\]

The derivation uses paraxial rays (rays close to the principal axis) and the principle of similar triangles.

Step 1: Set Up
Consider an object AB placed on the principal axis in front of a concave mirror with centre of curvature C and focus F. A ray from the top of the object (B) parallel to the principal axis strikes the mirror at D and reflects through F. Another ray from B passing through C strikes the mirror and retraces its path. These two reflected rays meet at B′, giving the image A′B′.

Step 2: Similar Triangles (first pair):
Triangles A′B′F and MPF are similar (where M is the point where the ray from B meets the mirror, and MP is the height of that ray above the axis):

Since PM = BA (same height, paraxial approximation):

Step 3: Similar Triangles (second pair):
Triangles A′B′A and ABP are also similar:

Step 4: Combining equations (1) and (2):

Step 5: Applying sign convention:

Using Cartesian sign convention:

• B′F = v − f
• FP = f
• A′P = v
• AP = u

Substituting:

Dividing through by v:

Rearranging:

Question: What happens if the lower half of a concave mirror is covered?

Simple steps:

• The obvious wrong guess: You might think only the upper half of the object will appear in the image.
• The correct reasoning: The laws of reflection still apply perfectly to every point on the remaining (upper) half of the mirror.
• Conclusion on image shape: Each point of the object still sends rays to the uncovered part of the mirror, so the full image of the whole object is still formed.
• Actual changes: Since the reflecting area is now half, fewer rays contribute to the image → the image becomes dimmer (half the intensity), not half the object.

Question: A mobile phone lies along the principal axis of a concave mirror. What does the image look like?

Simple steps:

• Understand the setup: The phone is lying flat along the principal axis, so different parts of the phone are at different distances from the mirror.
• Part perpendicular to the axis: The part of the phone on a plane perpendicular to the principal axis (say, point B) forms an image B′ in the same plane and of the same transverse size — so BC = B′C.
• Distortion occurs because the phone extends along the principal axis; different parts of it are at different object distances. The mirror equation yields different image distances and magnifications for each part → magnification is not uniform across the phone's length.
• Location matters: as you move the phone closer to or farther from the mirror, the degree of distortion changes because magnification varies across the phone's length.

Given: Radius of curvature R = 15 cm, so focal length f = −R/2 = −7.5 cm

Case (i): Object at u = −10 cm

Step 1: Write the mirror equation:

Step 2: Substitute:

Step 3: Solve for v:

Step 4: Find magnification:

Step 5: Interpret:

• v is negative → image is in front of mirror → real
• m is negative → image is inverted
• |m| = 3 → image is 3 times magnified

Case (ii): Object at u = −5 cm (between F and P)

Step 1: Substitute:

Step 2: Solve for v:

Step 3: Find magnification:

Step 4: Interpret:

• v is positive → image is behind the mirror → virtual
• m is positive → image is erect
• |m| = 3 → image is 3 times magnified

Given: Convex mirror → f = +R/2 = +1m. The jogger moves at 5 m/s towards the mirror.

Core idea: Use the rearranged mirror equation to find the image distance first, then calculate how fast the image shifts in 1 second.

Step 1: Find the image position at the starting distance.

Step 2: After 1 second, the jogger has moved 5 m closer. Find the new image position.

Step 3: The difference in the two image positions gives the image shift in 1 second = the average image speed.

Case (a): Jogger at u = −39 m

• At u = −39 m: v = \[\frac{1\times(-39)}{-39-1}=\frac{-39}{-40}=\frac{39}{40}\mathrm{~m}\]

After 1 s, u = −34 m:

Image shift:

Image speed ≈ 1/280 m/s.