34-4 Thin Lenses#

Prompts

Distinguish converging and diverging lenses. How do parallel rays behave? Which has a real focal point and which a virtual one?
For a converging lens: object inside vs outside the focal point—what type of image and what orientation?
For a diverging lens: what type of image can it form? Where is it located?
State the thin lens equation and the lens maker’s equation. What sign convention applies to \(r_1\) and \(r_2\)?
For a two-lens system: how do you find the final image? What if the first image lies past the second lens? How is overall magnification computed?

Lecture Notes#

Overview#

A thin lens has two spherical refracting surfaces with a common central axis. Light refracts at both surfaces.
Converging lens: parallel rays converge to a real focal point; \(f > 0\).
Diverging lens: parallel rays diverge; backward extensions meet at a virtual focal point; \(f < 0\).
Converging: real image if object outside \(F\); virtual if inside \(F\). Diverging: always virtual image.
Thin lens equation \(1/p + 1/i = 1/f\) (same form as mirrors); lens maker’s equation relates \(f\) to \(n\) and the radii.

Converging vs diverging lenses#

Type	Parallel rays	Focal point	\(f\)
Converging	Converge through a common point	Real	\(> 0\)
Diverging	Diverge; extensions meet at a point	Virtual	\(< 0\)

A converging lens is typically thicker at the center (e.g., double convex); a diverging lens is thinner at the center (e.g., double concave). Both refractions at the lens surfaces contribute to the net bending.

Images from converging lenses#

Object location	Image type	Image location	Orientation
Outside \(F\)	Real	Opposite side of lens	Inverted
Inside \(F\)	Virtual	Same side as object	Same as object

Images from diverging lenses#

A diverging lens forms only virtual images—always on the same side as the object, with the same orientation, and smaller. Regardless of object position, \(i < 0\).

Poll: Hidden lens—identify type and image

A lens is hidden behind a curtain. You are given three rays used to construct an image. What is the lens type and image type?

[FIGURE: Three rays (e.g., parallel→through focus, through focus→parallel, through center) either converging to a point or diverging with extensions meeting]

(A) Converging lens, real image
(B) Converging lens, virtual image
(C) Diverging lens, real image
(D) Diverging lens, virtual image

Poll: Another ray diagram—lens and image type

For the figure shown, determine the lens type and image type.

[FIGURE: Different ray diagram from the previous; design to test the other case (e.g., converging with object inside F → virtual)]

(A) Converging lens, real image
(B) Converging lens, virtual image
(C) Diverging lens, real image
(D) Diverging lens, virtual image

Poll: How can you see the image?

A lens forms an image of an arrow. Three rays from the arrowhead are drawn. The image in this sketch could be seen:

[FIGURE: Ray diagram showing either (a) converging rays to the right of the lens—real image—or (b) diverging rays with backward extensions meeting on the left—virtual image]

(A) By placing a screen at the image point
(B) Without a screen, by looking back through the lens
(C) By both techniques (A) and (B)
(D) Only if the lens is big enough
(E) None of these

Poll: Blocking half the lens

Jennifer blocks half the lens with a piece of paper, as shown. What happens to the image?

[FIGURE: Lens with lower half blocked by paper; object and image; rays from object through unblocked part]

(A) It disappears
(B) Only half of it is still seen
(C) It looks the same but gets slightly dimmer
(D) It gets fuzzy
(E) It depends on which part of the lens is blocked

Image location: lenses vs mirrors#

Important

For lenses (as for refracting surfaces): Real images form on the opposite side of the lens from the object; virtual images form on the same side. This is the reverse of mirrors.

Thin lens equation and magnification#

For paraxial rays:

(258)#\[ \frac{1}{p} + \frac{1}{i} = \frac{1}{f} \]

Sign convention: \(p > 0\); \(f > 0\) (converging), \(f < 0\) (diverging); \(i > 0\) (real image), \(i < 0\) (virtual image).

Lateral magnification (same as mirrors):

(259)#\[ m = -\frac{i}{p}, \qquad |m| = \frac{h'}{h} \]

Lens maker’s equation#

For a thin lens in air with index of refraction \(n\):

(260)#\[ \frac{1}{f} = (n - 1)\left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]

where \(r_1\) and \(r_2\) are the radii of curvature of the two surfaces. Use the refracting-surface sign convention: \(r > 0\) when the object faces a convex surface, \(r < 0\) when it faces a concave surface. (\(r_1\) is the surface nearer the object.)

If the lens is in a medium with index \(n_{\text{medium}}\), replace \(n\) with \(n/n_{\text{medium}}\).

Ray diagram: principal rays#

To locate an image, draw any two of these rays from an off-axis point:

Parallel to axis \(\to\) passes through \(F_2\) (or its extension for diverging).
Through \(F_1\) (or its extension) \(\to\) emerges parallel to axis.
Through center of lens \(\to\) emerges with no change in direction (lens is thin; surfaces nearly parallel at center).

For an object inside \(F_1\) of a converging lens, ray 2 uses the backward extension through \(F_1\).

Poll: Principal rays—real or virtual image?

The figure shows three principal rays for a converging lens. Which best describes the image?

[FIGURE: Converging lens with object; three principal rays either (a) converging to a point on the right—real—or (b) diverging with extensions meeting on the left—virtual]

(A) The image is real
(B) The image is virtual
(C) There is no image
(D) There are multiple images

Two-lens systems#

Step 1: Ignore lens 2; find image \(I_1\) produced by lens 1 using \(1/p_1 + 1/i_1 = 1/f_1\).

Step 2: Treat \(I_1\) as the object for lens 2, and find image \(I_2\) using \(1/p_2 + 1/i_2 = 1/f_2\). The object distance \(p_2\) is the distance from \(I_1\) to lens 2.

Caution

If \(I_1\) lies past lens 2 (to the right of it), then \(p_2 < 0\) when using the thin lens equation.

Overall magnification:

(261)#\[ M = m_1 \cdot m_2 \]

where \(m_1 = -i_1/p_1\), \(m_2 = -i_2/p_2\). \(M > 0\): final image same orientation as original object; \(M < 0\): inverted.

For two thin lenses in contact (or as a first approximation when close): \(1/f_{\text{tot}} = 1/f_1 + 1/f_2\) (combined power).

Example: Effective focal length of two adjacent lenses

Two converging lenses, each of focal length 50 cm, are placed 10 cm apart. For a distant object (at infinity), find the effective focal length of the composite system, assuming the effective “lens” is at the midpoint between them.

Solution: For adjacent lenses (or thin-lens approximation), \(1/f_{\text{tot}} = 1/f_1 + 1/f_2 = 1/50 + 1/50 = 2/50 = 1/25\) \(\Rightarrow\) \(f_{\text{tot}} = 25\) cm. (The 10 cm separation modifies this slightly; the adjacent-lens formula gives a first approximation.)

Example: Object and image from lens equation

An object is placed 20 cm from a converging lens of focal length +10 cm. (a) Draw a ray diagram. (b) Where is the image? (c) What is the magnification?

Solution: (a) Object outside \(F\); draw two principal rays. (b) \(1/i = 1/f - 1/p = 1/10 - 1/20 = 1/20\) \(\Rightarrow\) \(i = 20\) cm (real, on opposite side). (c) \(m = -i/p = -20/20 = -1\) (inverted, same size).

Summary#

Converging (\(f > 0\)): real focus; object outside \(F\) \(\to\) real inverted image; object inside \(F\) \(\to\) virtual upright image.
Diverging (\(f < 0\)): virtual focus; always virtual, upright, smaller image.
Thin lens equation: \(1/p + 1/i = 1/f\); magnification: \(m = -i/p\).
Lens maker’s equation: \(1/f = (n-1)(1/r_1 - 1/r_2)\).
Two-lens system: find \(I_1\), then \(I_2\); \(M = m_1 m_2\).