Mirrors, Lenses, and the Laws of Light
From Concave Mirrors to Converging Lenses — Master Every Ray Diagram
Light is the highest-weightage physics chapter in Class X, contributing 12–15 marks. Mirror formula, lens formula, and power of a lens are mandatory formula questions. Ray diagrams (concave mirror, convex lens, concave lens) are guaranteed 3–5 mark diagram questions in CBSE Boards. NTSE includes sign convention problems and nature/position/size of image MCQs.
Sign convention is the make-or-break skill — distances measured opposite to incident light are negative. For mirrors: u is always negative (real object); for concave f is negative. For lenses: f is positive for convex, negative for concave. Memorise all 6 object-position cases for both concave mirror and convex lens. Power in dioptres requires focal length in METRES. Time investment: 6–7 days.
Consider a slit of width \(a\). According to Huygens’ principle, every point on the slit acts as a secondary source of wavelets.
Similarly, higher order minima occur at: \[ a \sin \theta = n\lambda \]
A beam of monochromatic light passes through a narrow slit and spreads out. A student observes that decreasing slit width increases spreading.
Question: Explain the reason scientifically.
Answer: As slit width becomes comparable to wavelength, diffraction increases due to stronger wave interference effects.
Consider a spherical mirror of small aperture. Let an object AB be placed on the principal axis, and let its image A′B′ be formed after reflection. Using the geometry of the ray diagram, two pairs of triangles are obtained that are similar.
From the similar triangles formed near the pole of the mirror, we get a relation between the height of the object, the height of the image, and their distances from the pole:
\[ \frac{A'B'}{AB} = -\frac{v}{u} \]
By applying the second pair of similar triangles involving the focus F, we obtain another relation:
\[ \frac{A'B'}{AB} = -\frac{f-v}{f} \]
Equating the two expressions,
\[ -\frac{v}{u} = -\frac{f-v}{f} \]
On simplifying,
\[ \frac{f-v}{f} = \frac{v}{u} \]
\[ u(f-v)=fv \]
\[ uf-uv=fv \]
\[ uf=fv+uv \]
Dividing the whole equation by ufv, we get
\[ \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \]
Thus, the mirror formula is
\[ \boxed{\frac{1}{f}=\frac{1}{v}+\frac{1}{u}} \]
This relation holds for spherical mirrors when distances are measured according to the New Cartesian sign convention.
A driver uses a convex mirror. Explain why.
Answer: Convex mirrors provide wider field of view and always form erect images.
| Object Position | Image Position | Nature | Size |
|---|---|---|---|
| At infinity | At F | Real, inverted | Highly diminished |
| Beyond C | Between F and C | Real, inverted | Diminished |
| At C | At C | Real, inverted | Same size |
| Between C and F | Beyond C | Real, inverted | Magnified |
| At F | At infinity | Real, inverted | Highly enlarged |
| Between P and F | Behind mirror | Virtual, erect | Magnified |
| Object Position | Image Position | Nature | Size |
|---|---|---|---|
| Anywhere | Between P and F | Virtual, erect | Diminished |
A student places an object between F and P of a concave mirror.
Question: Describe the image formed.
Answer: Image is virtual, erect, and magnified behind the mirror.
A student draws only one ray while constructing a ray diagram.
Question: Why is the diagram incorrect?
Answer: At least two rays are needed to determine exact image position.
| Quantity | Concave Mirror | Convex Mirror |
|---|---|---|
| Focal length (f) | Negative | Positive |
| Radius (R) | Negative | Positive |
| Object distance (u) | Always Negative | Always Negative |
| Image distance (v) | Negative (real), Positive (virtual) | Always Positive |
| Image height (hᵢ) | Negative (inverted), Positive (erect) | Positive |
A student gets a positive image distance for a concave mirror in a numerical.
Question: What does this indicate?
Answer: Image is virtual and formed behind the mirror.
The mirror formula is derived using similar triangles formed in ray diagrams:
Equating and simplifying leads to:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
1/f = 1/v + 1/uA student gets positive magnification for a concave mirror.
Question: What does this imply?
Answer: Image is virtual and erect (object placed between P and F).
Corrected Geometry:
This gives refractive index of medium 2 with respect to medium 1.
Similarly:
\[ n_{12} = \frac{v_2}{v_1} \]
| Medium | Refractive Index |
|---|---|
| Air | ~1.0003 |
| Water | 1.33 |
| Glass | 1.5 |
| Diamond | 2.42 |
| Property | Optically Denser Medium | Optically Rarer Medium |
|---|---|---|
| Refractive Index | Higher | Lower |
| Speed of Light | Lower | Higher |
| Refraction | Towards normal | Away from normal |
| Object Position | Image Position | Size | Nature |
|---|---|---|---|
| At infinity | At F₂ | Highly diminished | Real, inverted |
| Beyond 2F₁ | Between F₂ & 2F₂ | Diminished | Real, inverted |
| At 2F₁ | At 2F₂ | Same size | Real, inverted |
| Between F₁ & 2F₁ | Beyond 2F₂ | Magnified | Real, inverted |
| Between F₁ & O | Same side | Magnified | Virtual, erect |
| Object Position | Image Position | Size | Nature |
|---|---|---|---|
| At infinity | At F₁ | Highly diminished | Virtual, erect |
| Anywhere | Between F₁ & O | Diminished | Virtual, erect |
| Feature | Convex Lens | Concave Lens |
|---|---|---|
| Shape | Thicker at center, thinner at edges | Thinner at center, thicker at edges |
| Curvature | Bulging outward (convex surfaces) | Curved inward (concave surfaces) |
| Action on light rays | Converges parallel rays to a focal point | Diverges parallel rays outward |
| Type of lens | Converging lens | Diverging lens |
| Focal length (sign) | Positive (f > 0) | Negative (f < 0) |
| Image type (in general) | Can form both real and virtual images | Always forms virtual images |
| Image orientation (typical) | Can be real and inverted or virtual and erect | Always erect |
| Image size (typical) | Can be magnified, same size, or diminished | Always diminished (smaller than object) |
| Common uses | Magnifying glass, cameras, telescopes, projectors, correcting hyperopia | Correcting myopia, peepholes, some laser‑beam expanders |
A student draws only one ray for a convex lens diagram.
Question: Why is it incorrect?
Answer: At least two rays are required to locate image accurately.
Academic Standards: ISO Optical Notation
Assign signs: \( u = -20 \), \( f = +10 \)
\[ \begin{aligned} \frac{1}{10} &= \frac{1}{v} - \frac{1}{-20}\\ \frac{1}{10} = \frac{1}{v} + \frac{1}{20}\\ \Rightarrow \frac{1}{v} &= \frac{1}{20}\\ \Rightarrow v &= +20 \, \mathrm{cm} \end{aligned} \]
A student obtains negative image distance for convex lens.
Question: What does it indicate?
Answer: Image is virtual and formed on same side as object.
Using ray‑diagram geometry and similar triangles formed by the object, image, and focal points, we can derive the lens formula. Consider a real object $AB$ and its real image $A'B'$ formed by a convex lens, with heights $h_o$ and $h_i$ respectively, object distance $u$, image distance $v$, and focal length $f$.
Equating the two expressions for $\frac{h_i}{h_o}$ yields: $$-\frac{v}{u} = \frac{v - f}{f}$$
On cross‑multiplying and simplifying: $$ \begin{aligned} -vf &= u(v - f) \\ -vf &= uv - uf \\ uf - vf &= uv \\ \end{aligned} $$
Dividing throughout by $uvf$: $$ \begin{aligned} \frac{uf}{uvf} - \frac{vf}{uvf} &= \frac{uv}{uvf} \\\\ \frac{1}{v} - \frac{1}{u} &= \frac{1}{f} \end{aligned} $$
Final Lens Formula: $$\boxed{\bbox[2pt]{\frac{1}{f} = \frac{1}{v} - \frac{1}{u}}}$$
Assign signs → \( u = -15 \), \( f = +10 \) \[ \begin{aligned} \frac{1}{10} &= \frac{1}{v} - \frac{1}{-15}\\\\ \frac{1}{10} &= \frac{1}{v} + \frac{1}{15}\\\\ \Rightarrow \frac{1}{v} &= \frac{1}{10} - \frac{1}{15} = \frac{1}{30}\\\\ v &= 30 \, \mathrm{cm} \end{aligned} \]
\[ \begin{aligned} m &= \frac{v}{u} \\\\&= \frac{30}{-15} \\\\&= -2 \end{aligned} \]
A student gets positive magnification for convex lens.
Question: What does it indicate?
Answer: Image is virtual and erect.
A lens has power −2 D. Identify the type of lens and its focal length.
Answer: Concave lens, \( f = -0.5 \, m \)
\[ \frac{1}{10} = \frac{1}{12} - \frac{1}{u} \]
\[ \frac{1}{u} = \frac{1}{12} - \frac{1}{10} \]
\[ \frac{1}{u} = \frac{5 - 6}{60} = -\frac{1}{60} \]
\[ u = -60 \, cm \]
\( u = -12 \, cm \)
\[ \begin{aligned} \frac{1}{f} &= \frac{1}{v} - \frac{1}{u}\\\\ \frac{1}{8} &= \frac{1}{v} + \frac{1}{12}\\\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{12} = \frac{1}{24}\\\\ v &= +24 \, \mathrm{cm} \end{aligned} \]
\[ \begin{aligned} h_i &= m \times h_o \\&= (-2)(2) \\&= -4 \, \mathrm{cm} \end{aligned} \]
\( u = -6 \, cm \)
\[ \begin{aligned} \frac{1}{8} &= \frac{1}{v} + \frac{1}{6}\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{6} \\ &= -\frac{1}{24}\\ v &= -24 \, \mathrm{cm} \end{aligned} \]
\[ \begin{aligned} h_i &= 4 \times 2 \\&= 8 \, \mathrm{cm} \end{aligned} \]
\( u = -24 \, cm \)
\[ \begin{aligned} \frac{1}{8} &= \frac{1}{v} + \frac{1}{24}\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{24} \\ &= \frac{2}{24} \\ &= \frac{1}{12}\\ \Rightarrow v &= +12 \, \mathrm{cm} \end{aligned} \]
\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{12}{-24} \\ &= -\frac{1}{2} \end{aligned} \]
\[ h_i = -\frac{1}{2} \times 3 = -1.5 \, \mathrm{cm} \]
Conclusion:
\( u = -3 \, cm \)
\[ \begin{aligned} \frac{1}{8} &= \frac{1}{v} + \frac{1}{3}\\ \frac{1}{v} &= \frac{1}{8} - \frac{1}{3} \\ &= -\frac{5}{24}\\ \Rightarrow v &= -4.8 \, \mathrm{cm} \end{aligned} \]
\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{-4.8}{-3} \\ &= 1.6 \end{aligned} \]
\[ \begin{aligned} h_i &= 1.6 \times 3 \\&= 4.8 \, \mathrm{cm} \end{aligned} \]
Conclusion:
Case (b) represents the working of a magnifying glass because:
\( u = -50 \, cm \)
\[ \begin{aligned} \frac{1}{20} &= \frac{1}{v} + \frac{1}{50}\\ \frac{1}{v} &= \frac{1}{20} - \frac{1}{50}\\ &= \frac{5 - 2}{100}\\ &= \frac{3}{100}\\ v &= \frac{100}{3} \\ &\approx 33.3 \, \mathrm{cm} \end{aligned} \]
\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{33.3}{-50} \\ &\approx -0.67 \end{aligned} \]
Conclusion:
\( u = -25 \, cm \)
\[ \begin{aligned} \frac{1}{20} &= \frac{1}{v} + \frac{1}{25}\\ \frac{1}{v} &= \frac{1}{20} - \frac{1}{25}\\ &= \frac{5 - 4}{100}\\ &= \frac{1}{100}\\ v &= 100 \, cm \end{aligned} \]
\[ \begin{aligned} m &= \frac{v}{u} \\ &= \frac{100}{-25} \\ &= -4 \end{aligned} \]
Conclusion:
Light travels in straight lines (rectilinear propagation). When light hits a smooth, polished surface it bounces back — this is reflection.
These laws hold for all types of reflecting surfaces — plane, concave, and convex.
Spherical mirrors are parts of a hollow sphere with a reflecting surface inside (concave) or outside (convex).
| Feature | Concave Mirror | Convex Mirror |
|---|---|---|
| Also called | Converging mirror | Diverging mirror |
| Reflecting surface | Inner (curved inward) | Outer (curved outward) |
| Focal point | Real, in front of mirror | Virtual, behind mirror |
| Image of distant object | Real, inverted, at F | Virtual, erect, diminished |
| Common uses | Shaving mirror, dentist's, headlights | Rear-view mirrors, security mirrors |
| Object Position | Image Position | Nature & Size |
|---|---|---|
| At infinity | At F | Real, inverted, highly diminished |
| Beyond C | Between F and C | Real, inverted, diminished |
| At C | At C | Real, inverted, same size |
| Between C and F | Beyond C | Real, inverted, magnified |
| At F | At infinity | Real, inverted, highly magnified |
| Between F and P | Behind mirror | Virtual, erect, magnified |
Interpreting magnification:
When light passes from one transparent medium to another, it changes its direction of travel. This bending is called refraction and happens because light travels at different speeds in different media.
When light goes denser → rarer medium: it bends away from normal (r > i).
When light goes rarer → denser medium: it bends toward normal (r < i).
| Medium | Refractive Index (approx.) |
|---|---|
| Vacuum / Air | 1.00 |
| Water | 1.33 |
| Crown Glass | 1.52 |
| Dense Glass | 1.65 |
| Diamond | 2.42 |
A higher refractive index means the medium is optically denser and light travels slower in it.
When light passes through a rectangular glass slab, it refracts at both surfaces. The emergent ray is parallel to the incident ray but laterally displaced (shifted sideways).
A lens is a transparent optical element bounded by two curved (or one curved, one plane) surfaces.
| Feature | Convex (Converging) Lens | Concave (Diverging) Lens |
|---|---|---|
| Centre | Thicker at centre | Thinner at centre |
| Effect on rays | Converges parallel rays to F | Diverges rays; appear from virtual F |
| Focal length | Positive (+) | Negative (−) |
| Power | Positive | Negative |
| Uses | Reading glasses, camera, magnifier | Myopia correction |
| Object Position | Image Position | Nature & Size |
|---|---|---|
| At infinity | At F₂ | Real, inverted, highly diminished |
| Beyond 2F₁ | Between F₂ and 2F₂ | Real, inverted, diminished |
| At 2F₁ | At 2F₂ | Real, inverted, same size |
| Between F₁ and 2F₁ | Beyond 2F₂ | Real, inverted, magnified |
| At F₁ | At infinity | Real, inverted, highly magnified |
| Between O and F₁ | Same side as object | Virtual, erect, magnified |
For a concave lens, image is always virtual, erect, and diminished, regardless of object position.
f is negative. u is always negative. v is negative for real image, positive for virtual image.
f is positive. u is always negative. v is always positive (virtual image always).
f is positive. u negative (object on left). v positive = real image (right side); v negative = virtual image (left side).
f is negative. Image always virtual, erect, diminished. v always negative (same side as object).
Enter any two of the three values. Leave one blank to solve for it.
Drag the object slider to change its position. See how the image changes in real time.
1/v + 1/u = 1/f (both positive sides, add). Lens → 1/v − 1/u = 1/f (subtract u side). The minus in lens formula trips students up most often.1/v + 1/u = 1/f. Lens: 1/v − 1/u = 1/f. These are different — confusing them is one of the most common exam errors.Get in Touch
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