Unit 06: Optics • Chapter 09 (Ray Optics)
Spherical Mirrors, Mirror Formula & Magnification
Cartesian sign conventions, paraxial ray approximation, geometrical derivation of $f = R/2$ and $1/v + 1/u = 1/f$, transverse & longitudinal magnification.
Kabhi kisi shiny stainless steel spoon (chammach) mein apna chehra dekha hai? Jab aap spoon ke andar wale hisse (concave cavity) mein dekhte hain, toh chehra ulta (inverted) dikhta hai, lekin jaise hi aap spoon ko aankh ke bilkul kareeb laate hain, achanak ek seedha aur vishalkay (magnified virtual) chehra nazar aane lagta hai! Wahi spoon jab ulti ki jaati hai (convex bulge), toh aap chahe kitni bhi door ya paas chale jayein, chehra hamesha chhota (diminished) aur seedha (erect) hi rehta hai (isiliye vehicles mein rear-view mirror convex hota hai taaki piche ka poora traffic ek chote mirror mein fit ho sake). Yeh saara magic kisi mystery ka nahi, balki do basic laws of reflection aur geometry ke Mirror Formula ($\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$) ka strict mathematical outcome hai!
1. Cartesian Sign Convention & Focus Relation
Strict rules for all optical derivations:
- All distances are measured from the Pole ($P$) along the principal axis.
- Distances measured in the direction of incident light are taken as Positive ($+$); opposite are Negative ($-$).
- Heights above principal axis are positive ($+$); below are negative ($-$).
2. Mirror Formula & Linear Magnification
Relation between object distance ($u$), image distance ($v$), and focal length ($f$):
Magnification sign rules: $m < 0 \implies$ Real & Inverted image; $m > 0 \implies$ Virtual & Erect image.
Relation Between Focal Length ($f$) and Radius of Curvature ($R$)
1. Consider a ray parallel to principal axis incident at point $M$ on a concave mirror of small aperture, reflecting through principal focus $F$. The normal to the surface at $M$ passes through the Center of Curvature $C$.
2. By law of reflection, angle of incidence equals angle of reflection: $\angle i = \angle r$. Since the incident ray is parallel to the principal axis, alternate interior angle $\angle MCF = \angle i = \theta$. Therefore, in $\triangle CFM$, $\angle CFM = 2\theta$ (exterior angle) and $\angle MCF = \theta$.
3. Drop perpendicular $MD$ from point $M$ to the principal axis. For paraxial rays (small aperture), point $D$ lies very close to pole $P$ ($D \approx P$), and $\tan\theta \approx \theta$, $\tan 2\theta \approx 2\theta$:
4. Since $2\theta = 2 \times \theta$:
The Mirror Equation ($\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$) for a Concave Mirror
Fig 9.1A: Real image formation by a concave mirror for an object placed beyond the Center of Curvature ($C$).
1. Consider similar right-angled triangles $\triangle A'B'P$ and $\triangle ABP$ (formed by rays incident at Pole $P$ with $\angle i = \angle r$):
2. Consider similar right-angled triangles $\triangle A'B'F$ and $\triangle MPF$ (where $MP \perp \text{axis}$ is the incident aperture height, so $MP = AB$):
3. Equating Eq. 1 and Eq. 2:
4. Applying Cartesian Sign Convention:
• Object distance $BP = -u$
• Image distance $B'P = -v$
• Focal length $FP = -f$
5. Dividing throughout by $v$ (or rearranging):
Linear Magnification: From Eq. 1 with sign conventions ($AB = +h_o$, $A'B' = -h_i$):
Trap 1: Mirror Immersed in Water or Another Medium
*Board Question:* A concave mirror of focal length $f = 20\text{ cm}$ in air is immersed in water ($\mu = 4/3$). What will be its new focal length?
*Correct Answer:* REMAINS UNCHANGED ($f = 20\text{ cm}$)!
Laws of reflection are independent of the surrounding refractive index ($f = R/2$ is a pure geometrical property). In contrast, a glass lens changes its focal length dramatically when immersed in liquid!
Trap 2: Covering Half the Aperture of a Mirror
*Exam Trap:* What happens to the image if the lower half of a concave mirror is painted black with opaque paint?
• The full complete image is still formed! Every single point on the mirror reflects rays from all parts of the object.
• However, the total amount of reflected light energy is halved; hence the brightness (intensity) of the image reduces by 50% ($I \propto \text{Area}$).
Trap 3: Convex Mirror Field of View
Why is a convex mirror preferred as a driver's rear-view mirror over a plane mirror?
1. It always forms an erect and diminished image ($m < 1$) for all positions of the object.
2. Because it curves outwards, it offers a much wider field of view than a plane mirror of the same size.
Trap 4: Longitudinal vs Transverse Magnification
If a small line element $du$ lies along the principal axis, differentiating $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$: $$-\frac{1}{v^2}dv - \frac{1}{u^2}du = 0 \implies m_L = \frac{dv}{du} = -\frac{v^2}{u^2} = -m^2$$ Longitudinal magnification of a small object along the axis is proportional to the square of linear transverse magnification ($m_L = -m^2$).
Problem: An object is placed at (i) $10\text{ cm}$, (ii) $5\text{ cm}$ in front of a concave mirror of radius of curvature $15\text{ cm}$. Find the position, nature, and magnification of the image in each case.
Solution: Focal length $f = R/2 = -15/2 = -7.5\text{ cm}$.
Case (i): $u = -10\text{ cm}$ (Object between $F$ and $C$):
Image is at $30\text{ cm}$ in front of mirror, Real, Inverted, and 3 times Magnified.
Case (ii): $u = -5\text{ cm}$ (Object between Pole and $F$):
Image is at $15\text{ cm}$ behind the mirror, Virtual, Erect, and 3 times Magnified.
Problem: A square wire of side $3.0\text{ cm}$ is placed $25\text{ cm}$ away from a concave mirror of focal length $10\text{ cm}$. What is the area enclosed by the image of the wire, given that the centre of the wire is on the principal axis and its plane is normal to the axis?
Solution:
1. Given: $u = -25\text{ cm}, f = -10\text{ cm}$.
2. Direct magnification formula:
3. Side of image of the square wire:
4. Area of image:
Mirror Formula
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
Focal Relation
$f = R/2$ (Medium Indep.)
Linear Magnification
$m = -\frac{v}{u} = \frac{f}{f-u}$
Areal Magnification
$A_{\text{image}} = m^2 A_{\text{obj}}$