The Mirror Equation

For an object on the principal axis of a spherical mirror, the image position follows from similar triangles in the ray diagram (object ray through C, ray to the pole). For paraxial rays the result is the mirror equation:

1v+1u=1f\boxed{\frac{1}{v} + \frac{1}{u} = \frac{1}{f}}

Derived for a real, inverted image in a concave mirror — but valid for every case (concave or convex, real or virtual image) once the Cartesian signs are used.

Ray diagrams and mirror equation cases for concave and convex mirrors

Lateral magnification:

m=h′h=−vu\boxed{m = \frac{h'}{h} = -\frac{v}{u}}

Sign decoding: m<0m < 0 — image inverted (and real, for mirrors); m>0m > 0 — image erect (and virtual). ∣m∣>1|m| > 1: magnified; ∣m∣<1|m| < 1: diminished.

The Case Map (Concave) and the One-Line Convex Story

Concave mirror (f negative), object moving in from infinity:

Object position Image Nature
Beyond C between F and C real, inverted, diminished
At C at C real, inverted, same size
Between C and F beyond C real, inverted, magnified
At F at infinity —
Between F and P behind the mirror virtual, erect, magnified (the make-up mirror!)

Convex mirror: for every real object position, the image is virtual, erect, diminished, located between P and F behind the mirror — which is why it gives a wide field of view (vehicle side-view mirrors), at the price of 'objects are closer than they appear'.

[JEE Tip] For numericals, skip the case map: substitute signed values in 1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f} and read the nature from the signs of v and m. The map is for ray-diagram questions and quick checks.

Solved Examples

Example 1: The half-covered mirror

The lower half of a concave mirror is covered with opaque material. What happens to the image of an object in front of it?

Solution:

  1. Every point of the remaining half still obeys the laws of reflection, and rays from every object point still reach it.
  2. The image remains complete — of the whole object.
  3. But the reflecting area is halved, so the image intensity drops (here to half). Covering a mirror (or lens) never crops the image; it only dims it.

Example 2: The phone along the axis

A mobile phone lies along the principal axis of a concave mirror. Why is its image distorted?

Solution:

  1. Different parts of the phone sit at different object distances u.
  2. Magnification m=−v/um = -v/u varies with u — each slice of the phone is magnified differently (the part on the plane perpendicular to the axis through C images at C, same size).
  3. Hence the image is distorted, and yes, the distortion depends on where the phone lies relative to the mirror.

Example 3: Two positions, two natures [NEET Numerical]

An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of R = 15 cm. Find the image position, nature and magnification.

Solution:

  1. f=−7.5f = -7.5 cm.
  2. (i) u=−10u = -10: 1v=1−7.5−1−10⇒v=−30\frac{1}{v} = \frac{1}{-7.5} - \frac{1}{-10} \Rightarrow v = -30 cm; m=−v/u=−3m = -v/u = -3. Real, inverted, magnified 3x, 30 cm in front.
  3. (ii) u=−5u = -5: 1v=1−7.5+15⇒v=+15\frac{1}{v} = \frac{1}{-7.5} + \frac{1}{5} \Rightarrow v = +15 cm; m=+3m = +3. Virtual, erect, magnified 3x, 15 cm behind — the object is inside F.

Example 4: The jogger in the mirror [JEE Numerical]

A jogger approaches a convex side-view mirror (R = 2 m) at 5 m/s. How fast does the image move when the jogger is 39 m, 29 m, 19 m and 9 m away?

Solution:

  1. f=+1f = +1 m. For u=−39u = -39: v=ufu−f=3940v = \frac{uf}{u-f} = \frac{39}{40} m. One second later (u=−34u = -34): v=3435v = \frac{34}{35} m.
  2. Image shift in that second: 3940−3435=1280\frac{39}{40} - \frac{34}{35} = \frac{1}{280} m — average speed 1280\frac{1}{280} m/s.
  3. Repeating: at 29 m → 1150\frac{1}{150} m/s; at 19 m → 160\frac{1}{60} m/s; at 9 m → 110\frac{1}{10} m/s.
  4. The image crawls when far and speeds up sharply as the jogger nears — exactly what you observe in a parked car.

Example 5: Convex mirror image hunt [NEET Numerical]

An object stands 10 cm in front of a convex mirror of f = +7.5 cm. Locate and describe the image.

Solution:

  1. 1v=1f−1u=17.5+110=730\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{7.5} + \frac{1}{10} = \frac{7}{30}.
  2. v=+307≈+4.3v = +\frac{30}{7} \approx +4.3 cm — behind the mirror.
  3. m=−v/u=+37≈0.43m = -v/u = +\frac{3}{7} \approx 0.43: virtual, erect, diminished — as always for a convex mirror.

Example 6: Where is the image the same size? [JEE Numerical]

For a concave mirror of f = -12 cm, where must the object stand for a real image of equal size?

Solution:

  1. Equal-size real image means m=−1m = -1, i.e. v=uv = u.
  2. The mirror equation: 2u=1f⇒u=2f=−24\frac{2}{u} = \frac{1}{f} \Rightarrow u = 2f = -24 cm.
  3. At the centre of curvature (24 cm in front) — object at C, image at C.

Example 7: Finding f from one image [JEE Numerical]

A concave mirror forms a real image 24 cm in front of it when the object stands 40 cm away (so u=−40u = -40 cm, v=−24v = -24 cm). Find f and m.

Solution:

  1. 1f=1v+1u=−124−140=−5+3120=−115\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = -\frac{1}{24} - \frac{1}{40} = -\frac{5+3}{120} = -\frac{1}{15}.
  2. f=−15f = -15 cm (R=−30R = -30 cm).
  3. m=−v/u=−2440=−0.6m = -v/u = -\frac{24}{40} = -0.6: real, inverted, diminished.

Example 8: Virtual-image magnification [NEET Numerical]

A make-up (concave) mirror of f = -15 cm shows your face erect and doubled in size. How far is your face?

Solution:

  1. Erect, doubled: m=+2=−v/u⇒v=−2um = +2 = -v/u \Rightarrow v = -2u.
  2. Mirror equation: 1−2u+1u=1−15⇒12u=−115\frac{1}{-2u} + \frac{1}{u} = \frac{1}{-15} \Rightarrow \frac{1}{2u} = -\frac{1}{15}.
  3. u=−7.5u = -7.5 cm — hold the mirror 7.5 cm away (inside F, as the case map demands).

Example 9: Why side-view mirrors are convex

Give the optical trade-off behind convex vehicle mirrors.

Solution:

  1. A convex mirror images the whole wide world as erect, diminished, virtual pictures between P and F — an enormous field of view in a small glass.
  2. The price: diminished images look farther than they are ('objects in mirror are closer than they appear').
  3. A plane mirror would show true distances but a far narrower field.

Example 10: One equation, every case

Verify the convex-mirror result of Example 5 satisfies the case map.

Solution:

  1. Computed: v=+4.3v = +4.3 cm (behind, between P and F since f=7.5f = 7.5 cm), m=+0.43m = +0.43.
  2. Case map for convex mirrors: image always virtual, erect, diminished, between P and F — all four boxes ticked.
  3. The algebra and the geometry always agree; use each to check the other in exams.