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Geometric Optics

6.1 Reflection and Refraction

Law of Reflection: The angle of incidence equals the angle of reflection: θi=θr\theta_i = \theta_r (both measured from the normal).

Snell”s Law: n1sinθ1=n2sinθ2n_1 \sin\theta_1 = n_2 \sin\theta_2.

Derivation of Snell’s law from Fermat’s principle. The optical path length from point AA in Medium 1 to point BB in medium 2 via a point on the interface at xx is:

Λ(x)=n1a2+x2+n2b2+(dx)2\Lambda(x) = n_1\sqrt{a^2 + x^2} + n_2\sqrt{b^2 + (d - x)^2}

Setting dΛ/dx=0d\Lambda/dx = 0:

n1xa2+x2=n2dxb2+(dx)2n_1 \frac{x}{\sqrt{a^2 + x^2}} = n_2 \frac{d - x}{\sqrt{b^2 + (d-x)^2}}

Which gives n1sinθ1=n2sinθ2n_1 \sin\theta_1 = n_2 \sin\theta_2. \blacksquare

6.2 Total Internal Reflection

When light travels from a denser to a rarer medium (n1>n2n_1 \gt n_2), total internal reflection occurs When θ1θc\theta_1 \geq \theta_c where:

sinθc=n2n1\sin\theta_c = \frac{n_2}{n_1}

Evanescent wave. Beyond the critical angle, the transmitted field decays exponentially:

EteκxE_t \propto e^{-\kappa x}

Where κ=2πλn12sin2θ1n22\kappa = \frac{2\pi}{\lambda}\sqrt{n_1^2 \sin^2\theta_1 - n_2^2}.

6.3 The Thin Lens Equation

1s+1s=1f\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}

Where ss is the object distance, ss' is the image distance, and ff is the focal length.

Sign convention (Cartesian): Distances are positive in the direction of light propagation. f>0f \gt 0 For converging lenses, f<0f \lt 0 for diverging.

Magnification:

M=ssM = -\frac{s'}{s}

Negative MM indicates an inverted image.

6.4 Lensmaker’s Equation

For a thin lens with radii of curvature R1R_1 and R2R_2:

1f=(n1)(1R11R2)\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

6.5 Matrix Optics (Ray Transfer Matrix)

A paraxial ray is described by a vector (yθ)\begin{pmatrix} y \\ \theta \end{pmatrix} where yy is the Height and θ\theta is the angle with the optical axis.

Free space propagation by distance dd:

Mprop=(1d01)M_{\mathrm{prop} = \begin{pmatrix} 1 & d \\ 0 & 1 \end{pmatrix}}

Thin lens of focal length ff:

Mlens=(101/f1)M_{\mathrm{lens} = \begin{pmatrix} 1 & 0 \\ -1/f & 1 \end{pmatrix}}

System matrix: The overall transformation is the product of individual matrices (applied in Reverse order): Msys=MnM2M1M_{\mathrm{sys} = M_n \cdots M_2 M_1}.

6.6 Mirror Equation

For a spherical mirror of radius RR (with R>0R \gt 0 for concave, R<0R \lt 0 for convex):

1s+1s=2R\frac{1}{s} + \frac{1}{s'} = \frac{2}{R}

The focal length is f=R/2f = R/2. The magnification is M=s/sM = -s'/s (negative for inverted images).

Derivation from the law of reflection. A ray from the object at height hh striking the mirror At height yy reflects such that θi=θr\theta_i = \theta_r. In the paraxial approximation (yRy \ll R), Applying the law of reflection and the small-angle approximation sinθθ\sin\theta \approx \theta:

hs+hs=2yR\frac{h}{s} + \frac{h'}{s'} = \frac{2y}{R}

Dividing through by yy and using h/s=y/sh/s = y/s, h/s=y/sh'/s' = y/s' (paraxial rays) yields the mirror Equation. \blacksquare

Worked Example: Concave mirror image formation

Problem. A concave mirror has radius of curvature R=40R = 40 cm. An object of height 2.0 cm is Placed 25 cm from the mirror. Find the image position, magnification, and nature.

Solution. f=R/2=20f = R/2 = 20 cm. Using 1/s+1/s=1/f1/s + 1/s' = 1/f: 1/s=1/201/25=(54)/100=1/1001/s' = 1/20 - 1/25 = (5 - 4)/100 = 1/100So s=100s' = 100 cm.

The image is real (positive ss') and on the same side as the object.

M=s/s=100/25=4.0M = -s'/s = -100/25 = -4.0.

Image height: h=Mh=4.0×2.0=8.0h' = Mh = -4.0 \times 2.0 = -8.0 cm (inverted, magnified by 4×\times).

6.7 Optical Instruments

Magnifying glass. Angular magnification when the image is at the near point DD:

M=1+DfM = 1 + \frac{D}{f}

Compound microscope. Total magnification:

Mtotal=LfoDfeM_{\mathrm{total} = -\frac{L}{f_o} \cdot \frac{D}{f_e}}

Where LL is the tube length, fof_o is the objective focal length, and fef_e is the eyepiece focal Length.

Refracting telescope. Angular magnification:

M=fofeM = -\frac{f_o}{f_e}

For large magnification, the objective should have a long focal length and the eyepiece a short one. The length of the telescope tube is approximately fo+fef_o + f_e.

Reflecting telescope. A concave primary mirror replaces the objective lens. The Cassegrain design Uses a secondary convex mirror to redirect the focus behind the primary. Advantages: no chromatic Aberration, lighter and cheaper for large apertures, and easier support structures.

Worked Example: Compound microscope magnification

Problem. A compound microscope has an objective with fo=4.0f_o = 4.0 mm and an eyepiece with fe=25f_e = 25 mm. The tube length is L=160L = 160 mm. Find the total magnification when the final image Is at the near point (D=250D = 250 mm).

Solution. Objective magnification: Mo=L/fo=160/4.0=40M_o = -L/f_o = -160/4.0 = -40.

Eyepiece magnification (image at near point): Me=1+D/fe=1+250/25=11M_e = 1 + D/f_e = 1 + 250/25 = 11.

Total magnification: Mtotal=Mo×Me=40×11=440M_{\mathrm{total} = M_o \times M_e = -40 \times 11 = -440}.

The negative sign indicates the image is inverted.