C/P 03 · Lenses, refraction and the eye
Light Through Glass and Lenses
Use the thin lens equation, Snell’s law and the critical angle to read optics measurements.
Passage
When light passes from one transparent material into another, its speed changes and its direction can bend. The index of refraction n of a material is the ratio of the speed of light in vacuum, c = 3 × 10⁸ m/s, to the speed of light in that material. Snell’s law relates the angles of the ray in the two materials: n₁ sin θ₁ = n₂ sin θ₂, where each angle is measured from the normal to the surface. Light going from a material of higher n to one of lower n bends away from the normal, and beyond a certain critical angle it cannot leave the first material at all.
A thin lens uses refraction at its curved surfaces to form images. For a lens of focal length f, an object at distance dₒ from the lens produces an image at distance dᵢ given by the thin lens equation. A positive dᵢ means a real image on the far side of the lens, and a negative dᵢ means a virtual image on the same side as the object. The magnification is m = −dᵢ/dₒ, where a negative m means the image is inverted.
Lenses and refraction are put to work in many instruments. A camera, a microscope and the eye all form images by bending light at curved surfaces, while optical fibers rely on a different effect: light stays trapped in a glass core because it strikes the boundary with the surrounding material at an angle that is too steep to escape. Both ideas are tested on a laboratory bench with a few simple measurements.
Part 1. A student uses an optical bench to study a thin converging lens whose focal length is not known. She places an illuminated object at several distances and slides a screen until a sharp image appears. Table 1 shows the object and image distances.
Table 1. Object and image distances for a thin converging lens.
| Trial | Object distance dₒ (cm) | Image distance dᵢ (cm) |
|---|---|---|
| 1 | 30 | 15 |
| 2 | 20 | 20 |
| 3 | 15 | 30 |
The student checks each image by moving the screen a few millimeters in each direction to confirm that the image is sharpest at the recorded position, and she measures distances from the center of the lens. She assumes that the lens is thin enough that its thickness can be neglected.
Part 2. The student then studies a ray of light inside a block of glass (n = 1.5) that strikes a flat glass–air surface from the inside. The air has n = 1.0. She increases the angle of incidence and records what the ray does at the surface. Table 2 shows the results.
Table 2. Behavior of a ray at a glass–air surface.
| Angle of incidence in glass | What happens at the surface |
|---|---|
| 30° | Ray refracts into the air, 49° from the normal |
| 40° | Ray refracts into the air, 75° from the normal |
| 50° | No ray enters the air; all light is reflected back into the glass |
In the human eye, the cornea and the lens together act like a single converging lens that must form a real image on the retina. If the eye is too long for its focusing power, or its focusing power is too strong, the image of a distant object forms in front of the retina. The student wonders whether the optics of the bench can explain such a vision problem.
Questions
Question 1 of 6
What is the focal length of the lens?
A focal length of 5 cm would give an image at 5 cm × 20 ÷ 15 for dₒ = 20 cm, not 20 cm. Check the data against 1/f = 1/dₒ + 1/dᵢ.
In Trial 2, dₒ = dᵢ = 20 cm. That happens when the object is at 2f, so f is half of 20 cm.
This is the object distance in Trial 1. It does not satisfy the thin lens equation with dᵢ = 15 cm.
Why the answer is what it is
In Trial 2, 1/f = 1/20 + 1/20 = 2/20, so f = 10 cm. The other trials agree: 1/30 + 1/15 = 3/30 = 1/10, and 1/15 + 1/30 = 3/30 = 1/10.
Review: Light & Optics · Equations: Waves, Optics & Modern Physics
Question 2 of 6
The student moves the object to 5 cm from the lens. Which description of the image is correct?
An object inside the focal length does not produce a real image. Solve 1/dᵢ = 1/10 − 1/5 and note the sign.
The object is closer than the focal length, so the rays leaving the lens spread apart and do not meet.
No image would form if the object were exactly at the focal point. At 5 cm, a virtual image does form.
Why the answer is what it is
1/dᵢ = 1/10 − 1/5 = −1/10, so dᵢ = −10 cm. The negative sign means the image is virtual and on the same side as the object. The magnification m = −(−10)/5 = +2 is positive, so the image is upright and twice as large.
Review: Light & Optics · Equations: Waves, Optics & Modern Physics
Question 3 of 6
Red light has a wavelength of 600 nm in air. When it enters the glass, its frequency and wavelength become:
The frequency of a wave is set by its source and does not change when it enters a new medium.
The frequency stays the same, and the speed decreases, so the wavelength must change.
Light slows down in glass, and since the frequency stays the same, the wavelength becomes shorter, not longer.
Why the answer is what it is
Frequency is set by the source and stays the same. The speed drops to c/n, so the wavelength drops by the same factor: 600 nm ÷ 1.5 = 400 nm.
Review: Light & Optics · Equations: Waves, Optics & Modern Physics
Question 4 of 6
What is the best estimate of the critical angle for the glass–air surface, based on Table 2?
The ray still refracts into the air at 30° and again at 40°, so the critical angle is greater than 40°.
At 40° the ray still enters the air, so total internal reflection has not begun yet.
At 50° all of the light is reflected, so the critical angle must be smaller than 50°.
Why the answer is what it is
The ray leaves the glass at 40° and is totally reflected at 50°, so the critical angle lies between them. This agrees with sin θc = n₂/n₁ = 1.0/1.5 ≈ 0.67, between sin 40° ≈ 0.64 and sin 50° ≈ 0.77.
Review: Light & Optics · Equations: Waves, Optics & Modern Physics
Question 5 of 6
The student replaces the air with water (n = 1.33) in contact with the glass. Compared with the glass–air surface, the critical angle will be:
Density is not what sets the critical angle. The ratio n₂/n₁ does.
The critical angle depends on both materials, since sin θc = n₂/n₁.
Total internal reflection is still possible because water has a lower index than glass.
Why the answer is what it is
The critical angle satisfies sin θc = n₂/n₁. With air this is 1.0/1.5 ≈ 0.67. With water it is 1.33/1.5 ≈ 0.89, which is a larger sine and therefore a larger angle.
Review: Light & Optics · Equations: Waves, Optics & Modern Physics
Question 6 of 6
A person’s eye focuses the image of a distant object in front of the retina. Which lens, placed in front of the eye, would move the image back onto the retina?
The eye already focuses too strongly. A converging lens would bring the rays together even sooner and move the image further forward.
A flat piece of glass does not change the convergence of the rays, so the image would stay where it is.
The image distance depends on the focal length of the whole optical system too, as the thin lens equation shows.
Why the answer is what it is
The image forms too close to the lens system, so the system is converging the light too strongly. A diverging lens spreads the rays slightly before they reach the eye and lowers the net focusing power, which moves the image back onto the retina. This is the correction used for nearsightedness.
Review: Vision & Optics · Light & Optics
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