Ray Optics Fundamentals
Concave Lens
Explore how diverging glass spreads light and forms virtual images. Experiment with correcting myopic eyes using eyeglasses, tracing diverging rays along an optics bench, and looking through a wide-angle door peephole.
Diverging Concave Lens Laboratory
Interact with the diverging glass elements to study refraction vectors, virtual focus, and real-world optics applications.
Live Telemetry
Lens Formula: 1/f = 1/v - 1/u- Active Setup
- Eyeglasses
- Focal Length (f)
- -10.0 cm
- Object Dist. (u)
- -25.0 cm
- Image Dist. (v)
- -7.1 cm
- Obj/Img Height
- 5.0 / 1.4 cm
- Magnification (m)
- +0.29x
What is a Concave Lens?
A **concave lens** (or **diverging lens**) is a piece of transparent optical material, typically glass or polymer, which is thinner in the center than at its peripheral boundaries. The surfaces of a concave lens curve inward, causing incoming parallel light rays to refract outward (diverge) upon transmission. As a result, the rays appear to emerge from a single virtual point in front of the lens, known as the **principal focus**.
Action of a Concave Lens: Refraction and Divergence
When light enters a medium with a higher refractive index (like glass), it slows down and bends toward the normal line. When exiting back into the air, it speeds up and bends away from the normal. Because a concave lens is thinner in the middle, the glass surfaces tilt outward as they move away from the center.
This geometry means that:
- Light passing exactly through the center (the **optical center**, \(O\)) travels along normal lines on both surfaces and passes straight through without refracting.
- Light striking the lens above or below the center is bent outward, away from the principal axis.
- Parallel light rays traveling parallel to the principal axis diverge completely. If you project these diverging rays backward as dotted lines, they intersect at a single point on the incident side, called the **principal focus** (\(F\)). The distance from \(O\) to \(F\) is the **focal length** (\(f\)).
Principal Ray Rules for a Concave Lens
To mathematically map or sketch the formation of images through a concave lens, we use three principal ray rules:
- **The Parallel Ray:** An incident ray parallel to the principal axis refracts outward, and its virtual backward extension passes through the principal focus (\(F\)) on the incident side.
- **The Central Ray:** An incident ray passing directly through the optical center (\(O\)) passes straight through without experiencing any angular deviation.
- **The Focal Ray:** An incident ray directed toward the principal focus (\(F\)') on the opposite side refracts parallel to the principal axis upon leaving the lens.
Summary of Image Formation Table
Unlike convex lenses, which can form real or virtual images of various sizes, a concave lens behaves very consistently. For any real object placed at any position relative to a concave lens, the image is **always virtual, upright, and diminished**.
| Object Position | Image Position | Nature | Size | Real-world Application |
|---|---|---|---|---|
| At Infinity | At Focus (\(F\)) | Virtual, Upright | Highly Diminished (Point) | Wide-angle telescope eyepiece, laser beam expanders |
| Beyond \(2F\) | Between \(F\) and \(O\) | Virtual, Upright | Diminished | Nearsighted eyeglasses, spyholes |
| At \(2F\) | Between \(F\) and \(O\) | Virtual, Upright | Diminished (Magnification \( < 0.5\)) | Optics instrumentation, viewfinder lenses |
| Between \(F\) and \(2F\) | Between \(F\) and \(O\) | Virtual, Upright | Diminished | Wide-field door viewer, optics lab rails |
| Between \(F\) and \(O\) | Between \(F\) and \(O\) | Virtual, Upright | Diminished (Forms very close to lens) | Corrective optics, specialized camera lenses |
Thin Lens Equation & Sign Convention
The mathematical relationship between focal length, object position, and image position is governed by the **Thin Lens Equation**:
The lateral **Magnification** (\(m\)) is calculated as:
Under standard **Cartesian Sign Convention**:
- The focal length (\(f\)) of a concave lens is always **negative**.
- The object distance (\(u\)) is **negative** when placed to the left (incident side) of the lens.
- The image distance (\(v\)) is **negative** because the image forms on the left (virtual, same side).
- The magnification (\(m\)) is **positive** (indicating an upright image) and always **less than 1.0** (indicating a diminished image).
Solved Examples
Example 1
An object of height 5.0 cm is placed at a distance of 15.0 cm from a concave lens of focal length 10.0 cm. Find the position, nature, and height of the image formed.
View Step-by-Step Solution
- Identify the parameters and apply sign conventions: Object height (hₒ) = +5.0 cm, Object distance (u) = -15.0 cm, Focal length (f) = -10.0 cm (always negative for concave lens).
- Recall the thin lens formula: 1/f = 1/v - 1/u.
- Rearrange to solve for 1/v: 1/v = 1/f + 1/u.
- Substitute the values: 1/v = 1/(-10.0) + 1/(-15.0) = -1/10 - 1/15.
- Find common denominator: 1/v = (-3 - 2) / 30 = -5/30 = -1/6.
- Calculate image distance (v): v = -6.0 cm.
- The negative image distance indicates a virtual, upright image on the same side of the lens as the object at a distance of 6.0 cm.
- Calculate magnification (m): m = v / u = -6.0 / (-15.0) = +0.4.
- Calculate image height (hᵢ): hᵢ = m * hₒ = 0.4 * 5.0 = 2.0 cm.
- The positive height confirms that the image is upright and diminished.
Final Answer: Image Distance (v) = -6.0 cm; Virtual, Upright, Diminished; Image Height (hᵢ) = +2.0 cm
Example 2
A concave lens has a focal length of 15.0 cm. At what distance should an object be placed in front of the lens so that it forms an image at 10.0 cm from the lens? Also, calculate its magnification.
View Step-by-Step Solution
- Identify parameters: Focal length (f) = -15.0 cm, Image distance (v) = -10.0 cm (concave lenses only form virtual images, which are on the same side, so v is negative).
- Recall the thin lens formula: 1/f = 1/v - 1/u.
- Rearrange to solve for 1/u: 1/u = 1/v - 1/f.
- Substitute the values: 1/u = 1/(-10.0) - 1/(-15.0) = -1/10 + 1/15.
- Find common denominator: 1/u = (-3 + 2) / 30 = -1/30.
- Calculate object distance (u): u = -30.0 cm. The object must be placed 30.0 cm in front of the lens.
- Calculate magnification (m): m = v / u = -10.0 / (-30.0) = +1/3 ≈ +0.33. The positive value confirms it is upright and reduced to one-third of the object size.
Final Answer: Object Distance (u) = -30.0 cm; Magnification (m) ≈ +0.33 (Upright, Diminished)
Example 3
A visitor stands at a distance of 40.0 cm from a door peephole containing a concave lens of focal length -8.0 cm. Where is their image formed, and what is the size reduction factor?
View Step-by-Step Solution
- Identify parameters: Object distance (u) = -40.0 cm, Focal length (f) = -8.0 cm.
- Recall the thin lens formula: 1/f = 1/v - 1/u.
- Rearrange to solve for 1/v: 1/v = 1/f + 1/u.
- Substitute values: 1/v = 1/(-8.0) + 1/(-40.0) = -1/8 - 1/40.
- Find common denominator: 1/v = (-5 - 1) / 40 = -6/40 = -3/20.
- Calculate image distance (v): v = -20/3 ≈ -6.67 cm.
- Calculate size reduction factor (magnification m): m = v / u = (-6.67) / (-40.0) = +0.167 (or 16.7% of actual height).
Final Answer: Image Distance (v) ≈ -6.67 cm; Size Reduction (m) ≈ 0.17x (Virtual, Upright, Diminished)
Self-Check Questions
Question 1
Why does a concave lens act as a diverging lens?
Show Answer & Explanation
A concave lens is thinner in the middle and thicker at the edges. When parallel rays pass through, the light bends toward the normal upon entering the glass and away from the normal upon exiting. Because of the curved shape curving inward on both sides, this double refraction causes all parallel rays to bend outward, diverging away from a single virtual focal point on the incident side.
Question 2
Why is the image formed by a concave lens always virtual, upright, and diminished?
Show Answer & Explanation
For a real object, one principal ray goes parallel to the axis and diverges away from the focus F, while another goes straight through the optical center. Since these refracted rays move away from each other on the transmission side, they can never intersect to form a real image. Instead, they intersect only when projected backward on the incident side. This intersection point always lies between the focus F and the lens, making the image virtual, upright, and smaller.
Question 3
How does a concave lens correct nearsightedness (myopia)?
Show Answer & Explanation
In a nearsighted (myopic) eye, the eyeball is too long or the lens is too strong, causing incoming parallel light from distant objects to focus in front of the retina. A concave eyeglass lens is placed in front of the eye to diverge the parallel light rays slightly before they enter the cornea. This pre-divergence shifts the final convergence point further back, allowing the eye's refracting system to focus the light directly onto the retina, restoring clear vision.