Light travels in straight lines in a homogeneous medium, and a thin beam is idealised as a ray. Ray optics treats light as rays whenever apertures are much larger than the wavelength, so diffraction can be ignored. The first phenomenon is reflection — the bouncing back of light from a polished surface.
The laws of reflection hold for every surface: (i) the incident ray, reflected ray and normal at the point of incidence are coplanar; (ii) the angle of incidence equals the angle of reflection, $\angle i = \angle r$, measured from the normal. A plane mirror forms a virtual, erect image of the same size, as far behind the mirror as the object is in front.
Spherical mirrors are portions of a reflecting sphere. A concave mirror reflects from the inner side; a convex mirror from the outer bulge. Key terms: the pole $P$, the centre of curvature $C$, the radius of curvature $R = PC$, the principal axis, and the principal focus $F$. For a mirror of small aperture the focus lies midway between pole and centre of curvature, giving $f = \dfrac{R}{2}$.
The image position follows from the mirror formula $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$, where $u$ is the object distance, $v$ the image distance and $f$ the focal length. The linear magnification is $m = \dfrac{h'}{h} = -\dfrac{v}{u}$: a negative $m$ means a real, inverted image; a positive $m$ a virtual, erect image. All distances obey the New Cartesian Sign Convention (measured from the pole; along the incident light positive, against it negative; heights up positive). Hence $f, R$ are negative for concave and positive for convex mirrors, and a real object has $u < 0$.
Image formation by a concave mirror: object beyond $C$ gives a real, inverted, diminished image between $F$ and $C$; at $C$, real, inverted, same size at $C$; between $C$ and $F$, real, inverted, magnified beyond $C$; at $F$, image at infinity; inside $F$, virtual, erect, magnified. A convex mirror always forms a virtual, erect, diminished image, giving a wide field of view — hence its use as a vehicle rear-view mirror.
An object is placed $15\,\text{cm}$ in front of a concave mirror of focal length $10\,\text{cm}$. Find the position, nature and magnification of the image.
Solution- Sign convention: $u = -15\,\text{cm}$, $f = -10\,\text{cm}$ (concave).
- Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-10} - \dfrac{1}{-15}$.
- $\dfrac{1}{v} = -\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{-3+2}{30} = -\dfrac{1}{30}$, so $v = -30\,\text{cm}$.
- Magnification $m = -\dfrac{v}{u} = -\dfrac{-30}{-15} = -2$.
Answer: Image is $30\,\text{cm}$ in front of the mirror, real, inverted and magnified two times ($m=-2$).
A convex mirror has a radius of curvature of $40\,\text{cm}$. An object is placed $30\,\text{cm}$ from it. Locate the image and state its nature.
Solution- For a convex mirror $f = +\dfrac{R}{2} = +\dfrac{40}{2} = +20\,\text{cm}$; object $u = -30\,\text{cm}$.
- $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{20} - \dfrac{1}{-30} = \dfrac{1}{20} + \dfrac{1}{30}$.
- $\dfrac{1}{v} = \dfrac{3+2}{60} = \dfrac{5}{60} = \dfrac{1}{12}$, so $v = +12\,\text{cm}$.
- $m = -\dfrac{v}{u} = -\dfrac{12}{-30} = +0.4$.
Answer: Image is $12\,\text{cm}$ behind the mirror, virtual, erect and diminished ($m = +0.4$).
Where must an object be placed before a concave mirror of focal length $20\,\text{cm}$ so that a real image of three times the object size is formed?
Solution- Real, inverted image means $m = -3 = -\dfrac{v}{u}$, hence $v = 3u$.
- $f = -20\,\text{cm}$. Substitute in $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$: $\dfrac{1}{3u} + \dfrac{1}{u} = \dfrac{1}{-20}$.
- $\dfrac{1 + 3}{3u} = -\dfrac{1}{20} \Rightarrow \dfrac{4}{3u} = -\dfrac{1}{20}$.
- $3u = -80 \Rightarrow u = -\dfrac{80}{3} \approx -26.7\,\text{cm}$.
Answer: The object must be placed about $26.7\,\text{cm}$ in front of the mirror (between $F$ and $C$).
An object of height $2\,\text{cm}$ is placed $30\,\text{cm}$ from a concave mirror of focal length $15\,\text{cm}$. Find the height and nature of the image.
Solution- $u = -30\,\text{cm}$, $f = -15\,\text{cm}$.
- $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-15} - \dfrac{1}{-30} = -\dfrac{2}{30} + \dfrac{1}{30} = -\dfrac{1}{30}$, so $v = -30\,\text{cm}$.
- $m = -\dfrac{v}{u} = -\dfrac{-30}{-30} = -1$.
- $h' = m \, h = -1 \times 2 = -2\,\text{cm}$.
Answer: Image is $2\,\text{cm}$ tall, real, inverted and the same size as the object, formed at the centre of curvature.
A concave mirror produces a virtual, erect image magnified $3$ times when the object is $10\,\text{cm}$ from it. Find the focal length of the mirror.
Solution- Virtual erect image: $m = +3 = -\dfrac{v}{u}$, so $v = -3u = -3(-10) = +30\,\text{cm}$ (behind mirror).
- $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{30} + \dfrac{1}{-10}$.
- $\dfrac{1}{f} = \dfrac{1}{30} - \dfrac{3}{30} = -\dfrac{2}{30} = -\dfrac{1}{15}$.
- $f = -15\,\text{cm}$.
Answer: The focal length is $15\,\text{cm}$ (negative sign confirms a concave mirror).
A $4.5\,\text{cm}$ object is placed $12\,\text{cm}$ in front of a convex mirror of focal length $15\,\text{cm}$. Find the image distance and size.
Solution- $u = -12\,\text{cm}$, $f = +15\,\text{cm}$ (convex).
- $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{15} - \dfrac{1}{-12} = \dfrac{1}{15} + \dfrac{1}{12}$.
- $\dfrac{1}{v} = \dfrac{4 + 5}{60} = \dfrac{9}{60}$, so $v = \dfrac{60}{9} \approx +6.7\,\text{cm}$.
- $m = -\dfrac{v}{u} = -\dfrac{6.7}{-12} = +0.56$; $h' = 0.56 \times 4.5 \approx 2.5\,\text{cm}$.
Answer: Image is about $6.7\,\text{cm}$ behind the mirror, virtual, erect and $2.5\,\text{cm}$ tall (diminished).