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Topic 13 of 15

Geometric Optics

Geometric optics treats light as rays that travel in straight lines, bending at interfaces according to Snell's law. This model accurately describes mirrors, lenses, and optical instruments whenever light wavelengths are much smaller than the objects involved.

Key Concepts

Law of Reflection

The angle of incidence equals the angle of reflection, both measured from the surface normal:

\theta_i = \theta_r

Snell's Law

When light crosses an interface between media with refractive indices

n_1

and

n_2

n_1\sin\theta_1 = n_2\sin\theta_2

Total Internal Reflection

When light travels from a denser medium to a less dense medium at an angle exceeding the critical angle

\theta_c = \arcsin(n_2/n_1)

, all light is reflected back.

Thin Lens Equation

Relates object distance

d_o

, image distance

d_i

, and focal length

f

1/f = 1/d_o + 1/d_i

. Converging lenses have

f > 0

, diverging lenses have

f < 0

Lateral Magnification

The ratio of image height to object height:

m = h_i/h_o = -d_i/d_o

. Negative

m

means an inverted image.

Mirror Equation

Same form as the thin lens equation:

1/f = 1/d_o + 1/d_i

. For concave mirrors

f = R/2 > 0

; for convex mirrors

f < 0

Key Equations

Snell's Law

n_1 \sin\theta_1 = n_2 \sin\theta_2

Relates angles of incidence and refraction at a boundary between two media.

Critical Angle

\sin\theta_c = \dfrac{n_2}{n_1}\quad(n_1 > n_2)

Minimum angle (from normal) for total internal reflection in the denser medium.

Thin Lens / Mirror Equation

\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}

Relates focal length to object and image distances.

Lateral Magnification

m = -\dfrac{d_i}{d_o}

Negative value means image is inverted relative to the object.

Speed of Light in Medium

v = \dfrac{c}{n}

Light slows down by factor $n$ (refractive index) in a medium.

Worked Example

Image Formed by a Converging Lens

Problem

A converging lens has focal length $f = 20\text{ cm}$ . An object is placed $d_o = 30\text{ cm}$ from the lens. Find the image distance and magnification.

Solution

Apply the thin lens equation.

\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}

So $d_i = 60\text{ cm}$ . The positive value means the image is real and on the far side of the lens.

Calculate the magnification.

m = -\frac{d_i}{d_o} = -\frac{60}{30} = -2

The image is real, inverted ( $m < 0$ ), and twice the size of the object.

Answer

d_i = 60\text{ cm}

m = -2

Practice

Exercises

7 problems

1 of 7

A ray of light travels from air ( $n=1.00$ ) into glass ( $n=1.52$ ) at an angle of incidence of $30.0°$ . What is the angle of refraction (in degrees)?

Your answer

2 of 7

Find the critical angle (in degrees) for total internal reflection at a glass–air interface where the glass has refractive index $n = 1.52$ .

Your answer

3 of 7

A converging lens of focal length $f = 15\text{ cm}$ has an object placed $40\text{ cm}$ from it. What is the image distance $d_i$ (in cm)?

Unlock Exercise 3