Convex Mirror — Properties, Ray Rules, Formula, Uses, Practice Problems & FAQs

Convex Mirror — Properties, Ray Rules, Formula, Uses, Practice Problems & FAQs

Convex Mirror — Properties, Ray Rules, Formula, Uses, Practice Problems & FAQs

A concise and original explanation of convex (diverging) mirrors: how they form images, the standard ray rules, mirror equation and magnification, practical applications, four fresh practice problems (with full solutions), and common questions answered.

Spherical mirrors — quick recap

A spherical mirror is a portion of a reflective spherical surface. Two standard types exist: concave (inner surface reflective — converging) and convex (outer surface reflective — diverging). In this article we focus on convex mirrors, which give a wide field of view and always form virtual images.

What is a convex mirror?

Imagine a hollow glass sphere coated with silver on the inside; when you look at the outside surface, that is a convex mirror. Because its reflecting surface bulges toward the incoming light, reflected rays spread (diverge) and the mirror is often called a diverging mirror.

Key characteristics of convex mirrors

  • Reflected rays diverge after reflection — they never meet in front of the mirror.
  • The image produced is always virtual, upright (erect), and diminished (smaller than the object).
  • As the object moves closer or farther, the virtual image moves along the line between the pole and the focus but remains behind the mirror.

Important terms

  • Pole (P): the geometrical centre of the mirror surface.
  • Centre of curvature (C): centre of the sphere of which the mirror is a part.
  • Principal axis: the straight line through P and C.
  • Focus (F): the point from which the reflected rays appear to diverge (for convex mirrors the focus is behind the mirror).
  • Radius of curvature (R): the radius of the parent sphere (distance PC).
  • Focal length (f): distance PF (for our sign convention we will take f positive for a convex mirror and measure u, v with signs explained below).

Ray-tracing rules (useful for drawing a ray diagram)

To locate the image of an object using a convex mirror, draw at least two of these standard rays from a point on the object:

  1. Parallel→apparent focus: A ray parallel to the principal axis reflects as if it originated from the focus behind the mirror.
  2. Toward focus→parallel: A ray aimed toward the focus (if extended behind the mirror) reflects and emerges parallel to the principal axis.
  3. Towards centre→back on itself: A ray directed toward the centre of curvature (behind the mirror) reflects back along the same path.
  4. Incidence at pole: A ray striking the pole reflects with the angle of incidence equal to the angle of reflection (like a plane mirror) and can be used as a check.

Image formation (qualitative)

Because reflected rays from a convex mirror diverge, they never form a real image in front of the mirror. If you extend the diverging rays backward (behind the mirror), they meet at a point that serves as the virtual image location. That image is always upright and smaller than the object, which is why convex mirrors are ideal where a wide field of view is desired.

Mirror formula & magnification

The mirror equation relates object distance \(u\), image distance \(v\), and focal length \(f\):

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

Magnification \(m\) is the ratio of image height \(h_i\) to object height \(h_o\). In terms of distances:

\[ m=\frac{h_i}{h_o}=-\frac{v}{u}. \]

Sign convention used here (consistent with the worked examples below): take the direction of incident light as positive. For a real object placed in front of the mirror we use \(u<0\); a virtual image located behind the mirror corresponds to \(v>0\); for convex mirrors we treat \(f>0\).

Common uses of convex mirrors

  • Vehicle side-mirrors (wide field of view and reduced blind-spots).
  • Safety mirrors at blind corners on roads and in stores to monitor aisles.
  • Entrance/security mirrors and decorative applications where a panoramic view is useful.

Practice problems (new numbers) — worked solutions

Q1. An object stands 30.0 cm in front of a mirror and a virtual image is observed 12.0 cm behind the mirror. Determine the type of mirror and its focal length.

Given (using our sign convention): \(u=-30.0\ \text{cm}\) and \(v=+12.0\ \text{cm}\).

Use the mirror formula:

\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{-30.0} + \frac{1}{12.0} = -\frac{1}{30} + \frac{1}{12}. \]

Compute (common denominator 60): \(-2/60 + 5/60 = 3/60 = 1/20\). So

\[ \frac{1}{f} = \frac{1}{20}\quad\Rightarrow\quad f = 20.0\ \text{cm}. \]

A positive focal length with the image behind the mirror indicates the mirror is convex.

Answer: Convex mirror, focal length \(f=20.0\ \text{cm}.\)

Q2. A convex mirror has focal length \(f=15.0\ \text{cm}\). An object is placed 28.0 cm in front of the mirror. Find the location of the image.

Given \(f=15.0\ \text{cm}\) and \(u=-28.0\ \text{cm}\). Solve for \(v\):

\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15.0} - \frac{1}{-28.0} = \frac{1}{15} + \frac{1}{28}. \]

Compute: common denominator 420 → \(28/420 + 15/420 = 43/420\). Thus

\[ \frac{1}{v} = \frac{43}{420}\quad\Rightarrow\quad v = \frac{420}{43}\approx 9.77\ \text{cm}. \]

Since \(v>0\), the image is virtual and located about \(9.77\ \text{cm}\) behind the mirror.

Answer: \(v \approx 9.8\ \text{cm}\) (virtual, behind mirror).

Q3. An object 12.0 cm tall produces a virtual image 4.0 cm tall in a convex mirror. Find the magnification of the mirror and state whether the image is erect or inverted.

Magnification \(m = h_i/h_o = 4.0/12.0 = 0.333\). Because the image height is positive (virtual & upright), the image is erect and diminished.

Answer: \(m = 0.333\) (image is upright / erect).

Q4. For a convex mirror the virtual image forms 20.0 cm behind the mirror when the object is 60.0 cm in front. Calculate the magnification.

Using distances \(v=+20.0\ \text{cm}\) and \(u=-60.0\ \text{cm}\), magnification:

\[ m = -\frac{v}{u} = -\frac{20.0}{-60.0} = \frac{20}{60} = 0.333. \]

Positive magnification indicates an upright virtual image, with size one third of the object.

Answer: \(m = 0.333\) (upright, diminished).

FAQs — quick answers

Q1. Why are convex mirrors used as vehicle side mirrors?
A: They provide a wider field of view and create a smaller, upright virtual image so the driver can see more area behind and to the side of the vehicle.

Q2. Does a convex mirror ever form a real image?
A: No — a convex mirror’s reflected rays always diverge, so any image is virtual (located by extending rays behind the mirror).

Q3. Why is magnification positive for convex mirrors?
A: The image produced by a convex mirror is upright relative to the object; with our sign convention this corresponds to a positive magnification \(m=h_i/h_o>0\).

Q4. How does magnification change as an object moves farther away?
A: For a convex mirror the virtual image gets closer to the focus behind the mirror and becomes progressively smaller; magnification decreases (image shrinks) as object distance increases.

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