AP Physics 2- 13.4 Images Formed by Lenses- Study Notes- New Syllabus
AP Physics 2- 13.4 Images Formed by Lenses – Study Notes
AP Physics 2- 13.4 Images Formed by Lenses – Study Notes – per latest Syllabus.
Key Concepts:
- Image Formation by a Lens
- Lens & Magnification Formula
- Ray Diagrams for Lenses
Image Formation by a Lens
Lenses form images by refraction of light rays. The type of image formed depends on whether the lens is convex (converging) or concave (diverging), and on the position of the object relative to the focal points of the lens.
Convex Lens (Converging Lens):
- Object beyond 2F: Image is real, inverted, and diminished, formed between F and 2F on the opposite side.
- Object at 2F: Image is real, inverted, and same size, formed at 2F on the opposite side.
- Object between F and 2F: Image is real, inverted, and magnified, formed beyond 2F.
- Object at F: Rays emerge parallel, image is formed at infinity.
- Object between F and lens: Image is virtual, erect, and magnified, formed on the same side as the object.
Concave Lens (Diverging Lens):
- For all object positions: Image is virtual, erect, and diminished, formed on the same side of the lens between the lens and its focus.
General Properties of Lens Images:
- Convex lenses can produce both real and virtual images.
- Concave lenses produce only virtual images.
- Real images are always inverted, while virtual images are always erect.
- The image size can be magnified, diminished, or the same size depending on object position.
Example :
A candle is placed between the focus and the optical center of a convex lens. Describe the nature, position, and orientation of the image formed.
▶️ Answer/Explanation
Since the object is between the focus (F) and the lens, the convex lens produces a virtual, erect, and magnified image.
The image appears on the same side of the lens as the object.
Example :
An object is placed at twice the focal length (2F) in front of a convex lens. Describe the characteristics of the image formed.
▶️ Answer/Explanation
When the object is at 2F, the image is formed at 2F on the opposite side of the lens.
The image is real, inverted, and same size as the object.
Example :
A student places an object in front of a concave lens. Describe the image formed regardless of the object position.
▶️ Answer/Explanation
A concave lens always produces an image that is virtual, erect, and diminished.
The image is formed between the lens and its focus, on the same side as the object.
Sign Conventions (Lens maker’s convention):
- All distances are measured from the optical center of the lens.
- Distances measured in the direction of the incident light are taken as positive.
- Distances measured opposite to the direction of incident light are negative.
- For a convex lens, the focal length \(\mathrm{f}\) is positive.
- For a concave lens, the focal length \(\mathrm{f}\) is negative.
Lens Formula
The relationship between the object distance, image distance, and focal length of a lens is given by the lens formula:
\(\mathrm{\dfrac{1}{f} = \dfrac{1}{v} – \dfrac{1}{u}}\)
- \(\mathrm{f}\) = focal length of the lens
- \(\mathrm{v}\) = image distance (measured from the optical center of the lens)
- \(\mathrm{u}\) = object distance (measured from the optical center of the lens)
Magnification by a Lens:
The magnification \(\mathrm{M}\) is the ratio of the height of the image to the height of the object:
\(\mathrm{M = \dfrac{h_i}{h_o} = \dfrac{v}{u}}\)
- If \(\mathrm{M > 0}\) → image is virtual and erect.
- If \(\mathrm{M < 0}\) → image is real and inverted.
Example
A convex lens has focal length \(\mathrm{f=+15\ cm}\). An object is placed 30 cm in front of the lens. Find the image distance \(\mathrm{v}\),
state whether the image is real or virtual, its orientation, and the magnification. (Assume light travels left→right; object is at the left of the lens so \(\mathrm{u=-30\ cm}\).)
▶️ Answer / Explanation
Step 1 — write lens formula:
\( \mathrm{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}} \)
Step 2 — solve for \(\mathrm{1/v}\):
Rearrange: \(\mathrm{\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}}\). Compute each term (digit-by-digit): $$ \mathrm{\dfrac{1}{f}=\dfrac{1}{15\ cm}=0.066\overline{6}\ cm^{-1}} $$ $$ \mathrm{\dfrac{1}{u}=\dfrac{1}{-30\ cm}=-0.033\overline{3}\ cm^{-1}} $$ Sum: $$ \mathrm{\dfrac{1}{v}=0.066\overline{6} + (-0.033\overline{3})=0.033\overline{3}\ cm^{-1}} $$
Step 3 — find \(\mathrm{v}\):
\( \mathrm{v=\dfrac{1}{0.033\overline{3}}=30\ cm} \)
Interpretation:
- \(\mathrm{v=+30\ cm}\) (positive) → image is formed on the opposite side of the lens from the object (right side) → real.
- Real images formed by lenses are inverted.
Step 4 — magnification:
\( \mathrm{M=\dfrac{v}{u}=\dfrac{30}{-30}=-1} \) So the image is the same size as the object (|M|=1) and inverted (negative sign).
Final answer: \(\mathrm{v=+30\ cm}\). Image is real, inverted, same size (magnification \(\mathrm{M=-1}\)).
Example
A concave lens has focal length \(\mathrm{f=-20\ cm}\). An object is placed 25 cm in front of the lens. Find the image distance \(\mathrm{v}\), state whether the image is real or virtual, its orientation, and the magnification. (Use \(\mathrm{u=-25\ cm}\).)
▶️ Answer / Explanation
Step 1 — lens formula:
\( \mathrm{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}} \quad\Rightarrow\quad \mathrm{\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}} \)
Step 2 — compute terms:
\( \mathrm{\dfrac{1}{f}=\dfrac{1}{-20\ cm}=-0.05\ cm^{-1}} \) \( \mathrm{\dfrac{1}{u}=\dfrac{1}{-25\ cm}=-0.04\ cm^{-1}} \) Sum: \( \mathrm{\dfrac{1}{v}=-0.05 + (-0.04) = -0.09\ cm^{-1}} \)
Step 3 — find \(\mathrm{v}\):
\( \mathrm{v=\dfrac{1}{-0.09} \approx -11.111\ cm \quad(\approx -11.11\ cm)} \)
Interpretation:
- \(\mathrm{v\approx -11.11\ cm}\) (negative) → image is on the same side of the lens as the object → virtual.
- Virtual images produced by diverging lenses are upright and diminished.
Step 4 — magnification:
\( \mathrm{M=\dfrac{v}{u}=\dfrac{-11.111}{-25}=+0.44444\ (\approx +0.44)} \) So image is upright (positive M), and its height is about 44% of the object height (diminished).
Optional numerical check (image height):
If object height \(\mathrm{h_o=4.0\ cm}\), image height \(\mathrm{h_i=M h_o=0.4444\times 4.0\approx 1.78\ cm}\).
Final answer: \(\mathrm{v\approx -11.11\ cm}\). Image is virtual, upright, diminished, with \(\mathrm{M\approx+0.44}\).
Ray Diagrams for Lenses
Ray diagrams are used to determine the location, type, size, and orientation of images formed by lenses.
Rules for Drawing Ray Diagrams (Convex Lens):
- A ray parallel to the principal axis passes through the focal point on the opposite side of the lens.
- A ray passing through the center of the lens continues straight without deviation.
- A ray passing through the focal point on the object side emerges parallel to the principal axis after refraction.
Rules for Drawing Ray Diagrams (Concave Lens):
- A ray parallel to the principal axis appears to diverge from the focal point on the same side of the lens.
- A ray directed toward the focal point on the opposite side emerges parallel to the principal axis.
- A ray passing through the center of the lens continues straight without deviation.
Image Characteristics Determined from Ray Diagrams:
- Location: Where the rays (or their extensions) intersect.
- Type: Real (rays actually meet) or virtual (extensions of rays appear to meet).
- Size: Compared to the object (magnified, same size, diminished).
- Orientation: Upright or inverted relative to the object.
Example :
A 4 cm tall object is placed 20 cm in front of a convex lens of focal length 10 cm. Use a ray diagram to describe the image properties.
▶️ Answer / Explanation
Step 1: Place object at 20 cm (2f) from lens.
Step 2: Ray diagram rules show that image forms at 20 cm on the other side of the lens.
Step 3: Image properties:
Location: At 20 cm beyond the lens
Type: Real (rays actually meet)
Orientation: Inverted
Size: Same size as the object (4 cm)