Chapter 9: Ray Optics and Optical Instruments

Ray optics (geometrical optics) treats light as straight-line rays. It works when the obstacle/aperture size is much larger than $\lambda$ (so diffraction is negligible). In this chapter we develop the mirror and lens equations, study refraction through plane, spherical, and prismatic surfaces, and apply these to optical instruments (eye, microscope, telescope).

Concept Map

• Reflection $\rightarrow$ Plane and spherical mirrors $\rightarrow$ Mirror formula $\rightarrow$ Magnification.
• Refraction $\rightarrow$ Snell's law $\rightarrow$ Slab (lateral shift) $\rightarrow$ Spherical surface ($n_2/v - n_1/u = (n_2-n_1)/R$) $\rightarrow$ Lens maker's formula $\rightarrow$ Lens formula $\rightarrow$ Power.
• Total internal reflection $\rightarrow$ Critical angle $\rightarrow$ Optical fibres, mirage, diamond brilliance.
• Prism $\rightarrow$ $A + D = i + e$ $\rightarrow$ Prism formula $\rightarrow$ Dispersion $\rightarrow$ Rayleigh scattering.
• Optical instruments $\rightarrow$ Eye (defects) $\rightarrow$ Simple/compound microscope $\rightarrow$ Refracting and reflecting telescopes.

Sign convention used throughout (Cartesian / "New Cartesian"):

• All distances are measured from the pole/optical centre.
• Distances measured in the direction of incident light are positive; those against it are negative.
• Heights measured above the principal axis are positive, below are negative.

9.1 Reflection by Spherical Mirrors

Definition

A spherical mirror is a portion of a reflecting sphere. The pole $P$ is the geometric centre of the mirror, centre of curvature $C$ is the centre of the sphere, principal axis is line $PC$, radius of curvature $R = PC$, and focal length $f = PF$, where $F$ is the principal focus.

For a concave mirror $R < 0$ and $f < 0$ (real focus in front). For a convex mirror $R > 0$ and $f > 0$ (virtual focus behind).

Law: angle of incidence equals angle of reflection, both measured from the local normal (which passes through $C$).

Derivation: $f = R/2$ for a paraxial concave mirror

Consider a paraxial ray $AB$ parallel to the principal axis striking the mirror at $B$ and reflecting through the focus $F$ on the axis. The normal at $B$ is $BC$. Let $\angle ABC = \theta$ (angle of incidence). Then $\angle CBF = \theta$ (angle of reflection).

Since $AB \parallel PC$, $\angle BCF = \theta$ (alternate angles). Triangle $BCF$ is therefore isosceles with $BF = CF$. For paraxial rays $B$ is close to $P$, so $BF \approx PF = f$ and $CF = CP - FP = R - f$. Hence

$f = R - f \quad\Rightarrow\quad f = \frac{R}{2}.$

Derivation: Mirror formula $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

Let $O$ be an axial object at distance $u$ from pole $P$, and $I$ the image at distance $v$. A ray from $O$ to $B$ reflects through $I$; the normal at $B$ passes through $C$.

In $\triangle OBC$: $\angle OBC = \alpha + \beta$ (exterior angle), where $\alpha = \angle BOP$, $\beta = \angle BCP$. In $\triangle OBI$: angle of incidence equals angle of reflection, so $\angle OBI = 2(\alpha + \beta)$. Also exterior angle of $\triangle OBI$ at $I$ gives $2(\alpha+\beta) = \alpha + \gamma$ where $\gamma = \angle BIP$. Hence

$\alpha + \gamma = 2\beta.$

For paraxial rays, $\alpha \approx BP/PO$, $\beta \approx BP/PC$, $\gamma \approx BP/PI$. Dividing by $BP$:

$\frac{1}{PO} + \frac{1}{PI} = \frac{2}{PC}.$

Applying the Cartesian sign convention with light travelling from left to right, $PO = -u$, $PI = -v$, $PC = -R$ (all in front of mirror):

$\frac{1}{-u} + \frac{1}{-v} = \frac{2}{-R} \;\Rightarrow\; \boxed{\;\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\;} \quad\text{with } f = \frac{R}{2}.$

Magnification

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

If $m > 0$ image is erect; if $m < 0$ inverted. $\vert m\vert > 1$ enlarged, $\vert m\vert < 1$ diminished.

Worked Example

A concave mirror has $f = -15\,\text{cm}$. An object is placed at $u = -20\,\text{cm}$. Find $v$ and $m$.

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{60}.$

So $v = -60\,\text{cm}$ (real, in front). $m = -v/u = -(-60)/(-20) = -3$. Image is real, inverted, magnified $3\times$.

Pitfalls

• Sign convention is non-negotiable; do not "memorise positive/negative" — derive it each time using "direction of incident light".
• $f$ of concave mirror is negative; calling it "positive" leads to wrong $v$.
• $m = -v/u$ (mirror) vs $m = v/u$ (lens) — students confuse the two.
• Behaviour of object between $F$ and $P$ for concave mirror: image virtual, erect, magnified.

9.2 Refraction at Plane Surfaces

Definition

When light passes obliquely from medium 1 (index $n_1$) to medium 2 (index $n_2$), it bends. Snell's law:

$n_1 \sin i = n_2 \sin r.$

Equivalent forms: $\dfrac{\sin i}{\sin r} = \dfrac{n_2}{n_1} = \dfrac{v_1}{v_2} = \dfrac{\lambda_1}{\lambda_2}$ (frequency stays constant across an interface).

Refractive index $n = c/v$ where $c$ is the speed of light in vacuum.

Derivation: Lateral shift through a glass slab

A ray enters a slab of thickness $t$ and index $n$ at angle $i$; inside, it bends to angle $r$ with $\sin i = n \sin r$. After traversing thickness $t$, it emerges parallel to the original direction (since the two faces are parallel), but laterally displaced by

$d = \frac{t \sin(i - r)}{\cos r}.$

Derivation. Inside the slab, the path along the ray has length $L = t/\cos r$. The perpendicular displacement of the emergent ray from the incident ray is $d = L \sin(i - r) = t \sin(i-r)/\cos r$.

For small $i$: $\sin(i-r) \approx i - r$, $\cos r \approx 1$, and $r \approx i/n$, so

$d \approx t\left(1 - \frac{1}{n}\right) i.$

Apparent depth of an object viewed from above through medium of index $n$:

$d_{\text{app}} = \frac{d_{\text{real}}}{n}.$

Worked Example

A coin lies at the bottom of a $12\,\text{cm}$ deep pool of water ($n = 4/3$). What is its apparent depth as seen from directly above?

$d_{\text{app}} = \frac{12}{4/3} = 9\,\text{cm}.$

Pitfalls

• Frequency $\nu$ does not change on refraction; wavelength does.
• For small-angle approximation, $d_{\text{app}} = d/n$ holds only near-normal viewing.
• A ray going from denser to rarer bends away from normal — the opposite direction from going rarer to denser.

9.3 Refraction at a Single Spherical Surface

Derivation: $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$

Consider a spherical refracting surface separating medium $n_1$ (left) from $n_2$ (right), pole $P$, centre of curvature $C$. An axial object $O$ at distance $u$ sends a paraxial ray to $B$ near $P$; the ray refracts and meets the axis at image $I$.

At $B$, the normal is $BC$. Let $\alpha = \angle BOP$, $\beta = \angle BCP$, $\gamma = \angle BIP$. Angle of incidence $i = \alpha + \beta$ (exterior angle of $\triangle OBC$). Angle of refraction $r = \beta - \gamma$ (exterior angle of $\triangle IBC$).

For small angles, Snell's law $n_1 \sin i = n_2 \sin r$ becomes $n_1 i = n_2 r$, i.e.

$n_1(\alpha + \beta) = n_2(\beta - \gamma).$

Using $\alpha \approx h/PO$, $\beta \approx h/PC$, $\gamma \approx h/PI$ for paraxial height $h$ and dividing by $h$:

$\frac{n_1}{PO} + \frac{n_2}{PI} = \frac{n_2 - n_1}{PC}.$

Applying Cartesian convention ($PO = -u$, $PI = +v$, $PC = +R$):

$\boxed{\;\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\;}.$

Worked Example

A fish lies $20\,\text{cm}$ below the surface of water ($n = 4/3$). Find the apparent depth viewed from above.

Treat the surface as flat ($R \to \infty$):

$\frac{n_2}{v} - \frac{n_1}{u} = 0,\quad v = \frac{n_2}{n_1} u.$

With $n_1 = 4/3$ (water), $n_2 = 1$ (air), $u = -20\,\text{cm}$: $v = (1)/(4/3) \cdot (-20) = -15\,\text{cm}$. Apparent depth $= 15\,\text{cm}$.

Pitfalls

• The formula has $n_2/v - n_1/u$, not $n_1/v - n_2/u$. The medium in which the image is formed sits on top of $v$.
• Sign of $R$: if $C$ lies on the side of refracted light, $R > 0$.

9.4 Thin Lens — Lens Maker's Formula and Lens Equation

Derivation: Lens maker's formula

A thin lens of material index $n$ in air has two surfaces with radii $R_1$ and $R_2$. Apply the spherical-surface formula at each surface.

Surface 1 (air $\to$ glass): refraction gives image $I_1$ at distance $v_1$ from the (effectively common) optical centre:

$\frac{n}{v_1} - \frac{1}{u} = \frac{n - 1}{R_1}.$

Surface 2 (glass $\to$ air): $I_1$ acts as object for the second surface. Refraction gives final image at $v$:

$\frac{1}{v} - \frac{n}{v_1} = \frac{1 - n}{R_2}.$

Adding:

$\frac{1}{v} - \frac{1}{u} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right).$

When the object is at infinity, $v = f$, giving the lens maker's formula:

$\boxed{\;\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\;}.$

If the lens sits in a medium of index $n_m$, replace $(n-1)$ by $(n/n_m - 1)$.

Thin lens formula and magnification

$\boxed{\;\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\;},\qquad m = \frac{v}{u} = \frac{h_i}{h_o}.$

For a converging lens $f > 0$; for diverging $f < 0$ (Cartesian).

Worked Example

A double-convex lens of glass ($n=1.5$) has $R_1 = +20\,\text{cm}$ and $R_2 = -20\,\text{cm}$. Find $f$.

$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \cdot \frac{2}{20} = 0.05\,\text{cm}^{-1}.$

So $f = 20\,\text{cm}$ (converging).

Pitfalls

• Lens-maker's formula derivation assumes a thin lens (both surfaces co-located). For a thick lens use full matrix optics.
• $R_2$ for a biconvex lens is negative (centre of curvature of second surface is to the left of pole).
• A glass lens in water has a longer focal length than in air because $(n_g/n_w - 1)$ is smaller than $(n_g - 1)$.

9.5 Power of a Lens and Combination of Lenses

Definition

Power $P = 1/f$ with $f$ in metres, unit dioptre (D). Converging lens has $P > 0$; diverging $P < 0$.

Combination in contact

Two thin lenses of focal lengths $f_1, f_2$ in contact. The image of the first is the object for the second. Adding the lens equations:

$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2},\qquad P_{\text{eq}} = P_1 + P_2.$

Magnification $m_{\text{eq}} = m_1 m_2$.

Combination separated by distance $d$

For two thin lenses separated by $d$,

$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2},\qquad P = P_1 + P_2 - d\,P_1 P_2.$

Worked Example

Two thin lenses of $P_1 = +5\,\text{D}$ and $P_2 = -2\,\text{D}$ are in contact. Find $f_{\text{eq}}$.

$P = 3\,\text{D} \Rightarrow f_{\text{eq}} = 1/3\,\text{m} \approx 33.3\,\text{cm}$ (converging).

Pitfalls

• Magnifications multiply, not add.
• Telescope/microscope objective + eyepiece are not in contact; use the separated-lens formula or treat them as a system with intermediate image.

9.6 Total Internal Reflection

Definition

When light travels from denser to rarer medium and $i > i_c$ (critical angle), it is totally reflected. By Snell with $r = 90^\circ$:

$\sin i_c = \frac{n_2}{n_1} = \frac{1}{n}\quad\text{(rarer = air)}.$

Conditions

1. Light must travel from denser to rarer medium.
2. $i > i_c$.

Applications

• Optical fibres: core ($n_1$) surrounded by cladding ($n_2 < n_1$); light propagates by repeated TIR with negligible loss.
• Brilliance of diamond: $n = 2.42$, $i_c \approx 24.4^\circ$; once light enters, most facets cause TIR.
• Mirage: hot road heats air at ground level reducing its index; light from sky bends progressively, undergoes TIR off the warm-air layer, giving an illusion of water.
• Prismatic binoculars/periscopes: $45^\circ$–$45^\circ$–$90^\circ$ prisms use TIR (since $i_c$ for glass $\approx 42^\circ < 45^\circ$).

Worked Example

For water $n = 4/3$. Find $i_c$.

$\sin i_c = 3/4 \Rightarrow i_c = 48.6^\circ$.

Pitfalls

• TIR does not occur going from rarer to denser.
• At $i = i_c$ exactly, refracted ray grazes the surface; intensity of reflected ray is high but not yet 100%.
• The cone of light escaping water from a point source has half-angle $i_c$.

9.7 Refraction Through a Prism

Definition

A prism has two refracting faces meeting at the refracting edge; the angle between them is the prism angle $A$. A ray bends towards the base on both refractions; the angle between emergent and incident rays produced is the deviation $\delta$.

Derivation: $A + \delta = i + e$ and Prism formula

Let $r_1, r_2$ be the refraction angles inside the prism at the two faces. In the quadrilateral formed by the two normals and the prism faces,

$r_1 + r_2 = A.$

At face 1, deviation $= i - r_1$; at face 2, $= e - r_2$. Total deviation:

$\delta = (i - r_1) + (e - r_2) = i + e - A \;\Rightarrow\; \boxed{i + e = A + \delta}.$

A plot of $\delta$ vs $i$ shows a single minimum $\delta = D_m$ where $i = e$ (by symmetry). Then $r_1 = r_2 = A/2$, and $i = (A+D_m)/2$. Applying Snell at face 1:

$\boxed{\;\mu = \frac{\sin\!\left(\dfrac{A+D_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}\;}.$

Thin prism (small $A$)

For small $A$, $\sin\theta \approx \theta$, so $D_m \approx (\mu - 1)A$. Deviation depends only on $\mu$ and $A$ (independent of $i$).

Worked Example

A prism of $A = 60^\circ$ and glass $\mu = 1.5$. Find $D_m$.

$\sin\frac{A+D_m}{2} = 1.5 \sin 30^\circ = 0.75 \Rightarrow \frac{A+D_m}{2} = 48.6^\circ.$

So $D_m = 2(48.6) - 60 = 37.2^\circ$.

Pitfalls

• The deviation is minimum (not zero) at $i = e$; do not equate $\delta = 0$.
• Thin-prism formula $D \approx (\mu - 1)A$ is independent of incidence angle, which is why prism-spectrometer formula tests are quick.

9.8 Dispersion Through a Prism

Definition

Different wavelengths have different $\mu$ (normal dispersion: $\mu_V > \mu_R$). After a prism, white light fans out into a spectrum.

Angular dispersion: $\theta = D_V - D_R = (\mu_V - \mu_R) A$ (thin prism).

Dispersive power:

$\omega = \frac{\mu_V - \mu_R}{\mu_Y - 1},$

where $\mu_Y$ is for the mean (yellow) wavelength.

Achromatic combination

Two thin prisms of materials with dispersive powers $\omega, \omega'$ and mean deviations $D, D'$ combined so net dispersion is zero but mean deviation is non-zero:

$\omega D + \omega' D' = 0.$

(The two prisms are oriented with bases opposite.) Conversely, a "direct-vision" prism gives dispersion without net deviation: $D + D' = 0$.

Pitfalls

• Dispersive power $\omega$ depends only on material, not on the prism angle.
• "$D = (\mu - 1)A$" is valid only for thin prisms; for $60^\circ$ prisms use the minimum-deviation formula.

9.9 Scattering of Light — Rayleigh's Law

For particles much smaller than $\lambda$, intensity of scattered light

$I_{\text{scat}} \propto \frac{1}{\lambda^4}.$

Consequences

• Blue sky: $\lambda_{\text{blue}} < \lambda_{\text{red}}$, so blue is scattered $\sim (700/450)^4 \approx 5.8$ times more than red.
• Reddish Sun at sunrise/sunset: light traverses a long atmospheric path; blue is scattered out, transmitted light is red-orange.
• White clouds: droplets are larger than $\lambda$, all colours scatter ~equally (Mie scattering), so clouds appear white.
• Danger signals (red): least scattered, travel furthest through fog.

Pitfalls

• Rayleigh works only for scatterers $\ll \lambda$; mist/clouds need Mie theory.

9.10 Optical Instruments

9.10.1 Human Eye

The eye is a converging-lens-and-retina system. Ciliary muscles change the lens shape to focus objects between the near point (least distance of distinct vision, $D = 25\,\text{cm}$) and far point (infinity for a normal eye) — this is accommodation.

Defects and corrections:

Worked Example — Myopia

A myopic patient cannot see beyond $50\,\text{cm}$. Find power of corrective lens.

A diverging lens must form a virtual image at $50\,\text{cm}$ of an object at infinity: $f = -50\,\text{cm} = -0.5\,\text{m}$, $P = -2\,\text{D}$.

9.10.2 Simple Microscope (Magnifier)

A single converging lens of small $f$ ($\lesssim 10\,\text{cm}$). The object is placed within the focal length so the virtual image lies at or beyond the near point.

Magnification with image at near point ($v = -D$):

$m = 1 + \frac{D}{f}.$

Magnification with image at infinity (relaxed eye):

$m = \frac{D}{f}.$

Derivation: visual angle subtended by object at near point is $\alpha_0 = h/D$; through lens it becomes $\alpha = h'/D$ (image at $D$) where $h' = h\cdot v/u$. Using lens formula with $v = -D$ and $u$ accordingly gives the result.

9.10.3 Compound Microscope

Two converging lenses: objective (very short $f_o$) and eyepiece ($f_e$, acts as simple magnifier). The objective forms a real, inverted, magnified image just inside the focal length of the eyepiece; the eyepiece then magnifies that image.

Let object lie just beyond $f_o$; image distance from objective $\approx L$ (tube length). Linear magnification of objective $m_o = -L/f_o$ (approx, image at infinity for eyepiece). Angular magnification of eyepiece $m_e = 1 + D/f_e$ (image at $D$) or $D/f_e$ (relaxed).

$\boxed{\;m = m_o \cdot m_e \approx -\frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)\;}\quad\text{(near-point)}.$

Relaxed-eye: $m \approx -L D/(f_o f_e)$.

9.10.4 Astronomical Telescope (Refracting)

Two converging lenses. Objective has large $f_o$ and large aperture; eyepiece has small $f_e$.

For a distant object, the objective forms a real image at its focal plane. The eyepiece, acting as a magnifier, examines that image.

Angular magnification, normal adjustment (final image at infinity, length $L = f_o + f_e$):

$m = -\frac{f_o}{f_e}.$

Near-point adjustment (final image at $D$, length $L = f_o + u_e$):

$m = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right).$

9.10.5 Reflecting Telescope (Cassegrain)

A large concave mirror replaces the objective. Magnification $m = f_o/f_e$ where $f_o = R/2$ of the mirror.

Advantages over refracting:

• No chromatic aberration (reflection independent of $\lambda$).
• Easier to make large mirrors than large lenses; only the front surface needs figuring.
• Mirror is supported from behind — no sagging under its own weight.
• Higher brightness and resolving power for given size.

Pitfalls

• For microscopes, $L$ is not simply the lens separation in the most general derivation; it is the image distance of the objective.
• $m = -f_o/f_e$ for a telescope, not $f_o + f_e$.
• A magnifier's $m = D/f$ is angular magnification (linear is much smaller).
• Reflecting telescopes are still subject to spherical aberration, fixed by using a paraboloidal mirror.

Solved Problems

1. A concave mirror of $f = -10\,\text{cm}$. An object is at $u = -15\,\text{cm}$. Find image distance, magnification, and nature.

$1/v = 1/f - 1/u = -1/10 + 1/15 = -1/30$, so $v = -30\,\text{cm}$. $m = -v/u = -30/15 \cdot \text{sign} = -2$. Image is real, inverted, magnified.

2. A glass slab of thickness $6\,\text{cm}$ ($n = 1.5$) lies between a coin and an observer's eye looking from directly above. By how much does the coin appear shifted?

Apparent shift $= t(1 - 1/n) = 6(1 - 2/3) = 2\,\text{cm}$ (towards observer).

3. A convex lens of $f = 20\,\text{cm}$ in air. An object is at $u = -30\,\text{cm}$. Find $v$, $m$.

$1/v = 1/f + 1/u = 1/20 + 1/(-30) = 1/60$. $v = +60\,\text{cm}$ (real, opposite side). $m = v/u = 60/(-30) = -2$.

4. Critical angle for glass ($n = 1.5$) in water ($n_w = 4/3$).

$\sin i_c = n_w/n_g = (4/3)/(1.5) = 8/9 \Rightarrow i_c = 62.7^\circ$.

5. A prism with $A = 60^\circ$ produces $D_m = 30^\circ$. Find $\mu$.

$\mu = \sin 45^\circ / \sin 30^\circ = (\sqrt 2/2)/(1/2) = \sqrt 2 \approx 1.414$.

6. A converging lens ($f_1 = +20\,\text{cm}$) and a diverging lens ($f_2 = -30\,\text{cm}$) are in contact. Find $f_{\text{eq}}$.

$1/f = 1/20 - 1/30 = 1/60$. $f_{\text{eq}} = 60\,\text{cm}$ (converging).

7. A compound microscope has $f_o = 1.0\,\text{cm}$, $f_e = 5\,\text{cm}$, tube length $L = 20\,\text{cm}$. Find magnification for near-point adjustment ($D = 25\,\text{cm}$).

$m \approx -(L/f_o)(1 + D/f_e) = -(20/1)(1 + 25/5) = -20 \cdot 6 = -120$.

JEE/NEET Edge Cases

• A silvered (semi-silvered) lens acts as a mirror with effective power $P_{\text{eff}} = 2 P_{\text{lens}} + P_{\text{mirror}}$ (light passes through the lens twice and reflects from the back-side mirror once).
• For a fish looking up out of water, the entire hemisphere of sky compresses into a cone of half-angle $i_c \approx 48.6^\circ$ — Snell's window.
• A convex lens in a medium of higher index becomes diverging.
• For two prisms in contact (achromatic combination): net dispersion zero but residual deviation gives chromatic-aberration-corrected lens (achromat).
• For a telescope, resolving power $\propto D/\lambda$ (Rayleigh): $\theta_{\min} \approx 1.22 \lambda/D$. Aperture matters more than magnification.
• The Cartesian sign convention for mirrors and lenses gives the same formula structure $1/v \pm 1/u = 1/f$; the $+$ in mirror eq vs $-$ in lens eq is purely from the directions of incident vs transmitted light.

Quick Recap

• $f_{\text{mirror}} = R/2$; mirror eq $1/v + 1/u = 1/f$; $m = -v/u$.
• Snell: $n_1 \sin i = n_2 \sin r$. Slab shift $t(1 - 1/n)$.
• Single surface: $n_2/v - n_1/u = (n_2 - n_1)/R$.
• Lens maker $1/f = (n-1)(1/R_1 - 1/R_2)$; lens eq $1/v - 1/u = 1/f$; $m = v/u$.
• Power $P = 1/f$ (D); in contact $P = \sum P_i$; separated by $d$: $P = P_1 + P_2 - dP_1P_2$.
• TIR: $\sin i_c = 1/n$ (denser$\to$rarer).
• Prism: $A + D = i + e$; $\mu = \sin((A+D_m)/2)/\sin(A/2)$; thin: $D = (\mu-1)A$.
• Rayleigh: $I \propto 1/\lambda^4$.
• Magnifier: $m = 1 + D/f$ (near pt), $D/f$ (relaxed).
• Compound microscope: $m \approx -(L/f_o)(1 + D/f_e)$.
• Telescope: $m = -f_o/f_e$ (normal); reflecting type avoids chromatic aberration.

Formula Sheet