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JEE Mains Physics · Ray Optics

Plane and Spherical Mirrors

A plane mirror forms an erect, same-size image as far behind it as the object is in front; a spherical mirror obeys 1/v + 1/u = 1/f with f = R/2 and m = −v/u, every distance measured from the pole.

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Why this matters

Twenty-nine PYQs, twenty-one of them multiple choice, and four from 2026. Eight are about plane mirrors: the nature of the image, the deviation of a reflected ray, a mirror moved towards the object, images between two mirrors and the law of reflection in vector form. Sixteen use the mirror formula; five of those give the distance between object and image together with the magnification. Five follow a moving object, or a rod lying along the axis.

Concept 1 of 3: Reflection at a plane mirror

A plane mirror makes an image as far behind it as the object is in front, on the same normal. The image is virtual, erect and the same size, but left and right are swapped. Each reflected ray turns through 180° − 2i. Because the image sits at twice the mirror's distance from a fixed object, moving the mirror moves the image twice as far.

Definition

  • Image: the same distance behind the mirror, the same size, virtual, erect and laterally inverted (m=+1m = +1).
  • Deviation of a ray reflected at an angle of incidence ii: δ=180∘−2i\delta = 180^{\circ} - 2i.
  • Object fixed, mirror moved by dd along its normal: the image moves 2d2d the same way. Mirror fixed, object moved by dd: the image moves dd. A mirror turned through θ\theta turns the reflected ray through 2θ2\theta.
  • Vector form: with unit incident direction a^\hat a and unit normal n^\hat n, the reflected direction is b^=a^−2(a^⋅n^)n^\hat b = \hat a - 2(\hat a\cdot\hat n)\hat n. The part along the mirror is kept; the part along the normal is reversed.
  • Two mirrors at an angle θ\theta: 360∘θ−1\dfrac{360^{\circ}}{\theta} - 1 images when 360∘/θ360^{\circ}/\theta is even (three at 90∘90^{\circ}). A ray reflected once from each mirror turns through 360∘−2θ360^{\circ} - 2\theta in all.
  • Parallel mirrors give an endless row of images. Build it in steps: every image formed in one mirror is an object for the other.
  • To see an image in a mirror, the eye must lie on a line from the image through the mirror. Lines from the image through the two edges of the mirror bound the region where it can be seen.

Reflection at a plane mirror

δ=180∘−2i,b^=a^−2(a^⋅n^)n^,n=360∘θ−1\delta = 180^{\circ} - 2i, \qquad \hat b = \hat a - 2(\hat a\cdot\hat n)\hat n, \qquad n = \frac{360^{\circ}}{\theta} - 1

Worked example

An object stands 15 cm in front of a plane mirror. (a) The mirror is moved 5 cm farther from the object, parallel to itself. How far does the image move? (b) A ray meets the mirror at an angle of incidence of 25∘25^{\circ}. Through what angle is it deviated?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q17Moderate

Example 1 · Ray Optics · Plane and Spherical Mirrors

An object is placed at a distance of 12 cm12\text{ }cm in front of a plane mirror. The virtual and erect image is formed by the mirror. Now the mirror is moved by 4 cm4\text{ }cm towards the stationary object. The distance by which the position of image would be shifted, will be:

Deviation is not the angle of reflection

A ray reflected at 35° to the normal is turned through 180° − 70° = 110°, not 35° or 70°. Deviation is measured from the ray's original direction.

Moving the mirror is not moving the object

With the object fixed, a mirror moved by d shifts the image by 2d. With the mirror fixed, an object moved by d shifts the image by d. Read which one moves.

A plane-mirror image is erect

Lateral inversion swaps left and right; it does not turn the image upside down. The image is virtual, erect and the same size, so m = +1.
Drill 7 more on reflection at a plane mirror

Concept 2 of 3: The mirror formula and magnification

Every spherical-mirror question uses one equation, 1/v + 1/u = 1/f, and one sign convention. Measure every distance from the pole, positive in the direction the incident light travels and negative against it. A real object in front of the mirror then has u < 0; a concave mirror has f < 0 and a convex mirror f > 0. A real image forms in front (v < 0) and is inverted; a virtual image forms behind (v > 0) and is erect.

Definition

  • Sign convention (Cartesian, used on every page of this chapter): distances from the pole or the optical centre; positive along the incident light; heights positive upward.
  • f=R/2f = R/2. Concave: f=−∣R∣/2f = -|R|/2. Convex: f=+∣R∣/2f = +|R|/2. A mirror works by reflection only, so its focal length is the same in air and in a liquid.
  • 1v+1u=1f\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}, m=−vu=ff−um = -\dfrac{v}{u} = \dfrac{f}{f - u}.
  • Concave mirror, object beyond C: real, inverted, diminished. Between F and C: real, inverted, magnified, beyond C. Inside F: virtual, erect, magnified, behind the mirror. At F: image at infinity.
  • Convex mirror: always virtual, erect and diminished, between the pole and F. A positive magnification below 1 with a real object means a convex mirror.
  • Given m and the object-image distance: write v=−muv = -mu, then set ∣v−u∣|v - u| equal to the given distance. A real image lies on the object's side; a virtual one lies behind the mirror, so the two distances add.
  • The same size of image at two object positions: one image is real (m=−km = -k), the other virtual (m=+km = +k).
  • Distances x1x_1, x2x_2 of object and image from the focus satisfy x1x2=f2x_1 x_2 = f^{2} (Newton's form).
  • To find f by parallax the image must be real, so the object goes anywhere beyond F.

Mirror formula

1v+1u=1f,f=R2,m=−vu=ff−u\frac{1}{v} + \frac{1}{u} = \frac{1}{f}, \qquad f = \frac{R}{2}, \qquad m = -\frac{v}{u} = \frac{f}{f - u}

Worked example

A concave mirror forms a real image twice the size of the object, and the object and image are 18 cm apart. (a) Find the focal length. (b) Where must the object be placed to get an erect image of the same size?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 January 2025 · Q88Moderate

Example 2 · Ray Optics · Plane and Spherical Mirrors

A concave mirror produces an image of an object such that the distance between the object and image is 20 cm . If the magnification of the image is ' -3 ', then the magnitude of the radius of curvature of the mirror is :

m = −v/u for a mirror, v/u for a lens

The mirror formula has a plus sign and its magnification a minus sign. Using the lens form m = v/u for a mirror makes every real image come out erect.

An erect, smaller image means a convex mirror

For a real object, a concave mirror never gives an erect diminished image. An erect image smaller than the object can come only from a convex mirror, with f > 0.

A mirror's focal length does not depend on the medium

f = R/2 contains no refractive index. Dipping a mirror in water changes nothing; dipping a lens in water does.

Two positions, two kinds of image

When the same image size appears at two object positions, one image is real and one virtual. Use m = −k for one and m = +k for the other, never the same sign twice.
Drill 15 more on the mirror formula and magnification

Concept 3 of 3: Image speed for a moving object

Differentiate the mirror formula with f fixed and the image's speed follows from the object's. Along the axis the image speed is m² times the object speed, where m is the magnification at that instant. Across the axis it is m times. A long rod lying along the axis is not a small object: image each end on its own.

Definition

  • Differentiating 1v+1u=1f\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} with f fixed: dvdt=−v2u2dudt=−m2dudt\dfrac{dv}{dt} = -\dfrac{v^{2}}{u^{2}}\dfrac{du}{dt} = -m^{2}\dfrac{du}{dt}.
  • So along the axis the image speed is m2m^{2} times the object speed, with m=ff−um = \dfrac{f}{f - u} at that instant. Across the axis, the image speed is m times the object speed.
  • A rod lying along the axis: find the image of each end with the mirror formula and subtract. Only a SHORT rod has an image length of m2m^{2} times its own.
  • An object moving at constant speed: find its new u at the given time, then use the mirror formula again.
  • Acceleration of the image: write v as a function of u, then differentiate twice with respect to time.
  • A mirror on a moving car: use the speed of the other car relative to the mirror as du/dtdu/dt.

Image speed along the axis

dvdt=−m2dudt,m=ff−u\frac{dv}{dt} = -m^{2}\frac{du}{dt}, \qquad m = \frac{f}{f - u}

Worked example

A convex mirror of focal length 2 m is fixed on a parked bike. A car approaches from behind at 15 m/s. How fast does the car's image move when the car is 8 m from the mirror?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 18 · Q3Moderate

Example 3 · Ray Optics · Plane and Spherical Mirrors

Car B overtakes another car A at a relative speed of 40 ms−140\text{ }ms^{-1}. How fast will the image of car B appear to move in the mirror of focal length 10 cm10\text{ }cm fitted in car A, when the car B is 1.9 m1.9\text{ }m away from car A?

Along the axis, speed scales with m², not m

Differentiating the mirror formula gives dv/dt = −m² du/dt. Using m alone gives an image speed that is too large for a diminishing convex mirror.

A long rod is not a short object

The rule that the image length is m² times the rod's length holds only when the rod is short compared with its distance from F. For a long rod, image each end.

Use the speed relative to the mirror

When the mirror rides on one car and the object is another car, du/dt is the rate at which the gap closes: their relative speed.
Drill 4 more on image speed for a moving object

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Reflection at a plane mirror

    Reflection at a plane mirror

    δ=180∘−2i,b^=a^−2(a^⋅n^)n^,n=360∘θ−1\delta = 180^{\circ} - 2i, \qquad \hat b = \hat a - 2(\hat a\cdot\hat n)\hat n, \qquad n = \frac{360^{\circ}}{\theta} - 1
  • The mirror formula and magnification

    Mirror formula

    1v+1u=1f,f=R2,m=−vu=ff−u\frac{1}{v} + \frac{1}{u} = \frac{1}{f}, \qquad f = \frac{R}{2}, \qquad m = -\frac{v}{u} = \frac{f}{f - u}
  • Image speed for a moving object

    Image speed along the axis

    dvdt=−m2dudt,m=ff−u\frac{dv}{dt} = -m^{2}\frac{du}{dt}, \qquad m = \frac{f}{f - u}

Watch out for (10)

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