The Human Eye and the Colourful World – Numerical Problems with Stepwise Solutions
20 Numerical Problems with Step‑by‑Step Solutions
CBSE Board Examination Focus:
- Lens formula, magnification and power calculations.
- Spectacle prescription problems and practical focal length measurements.
- Numerical applications of dispersion and rainbow geometry where applicable.
Content Bank — Important Formulae
- Lens formula:
1/f = 1/v - 1/uwhere f, v, u are in same units. Use<sub>and<sup>where needed. - Magnification:
m = v/u = h_i/h_o(negative sign indicates inversion). - Power of a lens:
P = 1/f (in metres), unit: dioptre (D). - Conversion:
1 m = 100 cm; 1 cm = 0.01 m. - Sign convention (Cartesian): real object <em>u</em> = negative, real image <em>v</em> = positive, convex lens <em>f</em> = positive.
- Approx. rainbow geometry: primary bow angle ≈
42°for red.
Numerical Problems — Topic wise
1. Lens formula & image distance
An object is placed 30 cm in front of a convex lens of focal length 10 cm. Find the image distance and describe the image.
- Given:
u = -30 cm(real object; negative by Cartesian sign convention),f = +10 cm. - Use lens formula:
1/f = 1/v - 1/u. - Substitute:
1/10 = 1/v - 1/(-30) = 1/v + 1/30. - Solve for
1/v:1/v = 1/10 - 1/30 = (3-1)/30 = 2/30 = 1/15. - So,
v = 15 cm. Image is real, inverted and formed 15 cm on the other side of the lens.
2. Magnification calculation
Using results from Problem 1 (object 30 cm, image 15 cm), calculate linear magnification and state whether image is magnified or diminished.
- Magnification
m = v/u. Herev = +15 cm,u = -30 cm. - So
m = 15/(-30) = -1/2 = -0.5. - Negative sign → image is inverted. Magnitude 0.5 < 1 → image is diminished (half the object height).
3. Focal length from image-object separation
An object and its real image are 60 cm apart when using a convex lens. If the image is real and inverted and the lens is thin, find the focal length.
- Let object distance be
uand image distance bev. Givenv - u = 60 cmif object is on one side and image on the other; more generally|v| + |u| = 60. - For a real image by convex lens, both distances are positive in magnitude. Use lens formula
1/f = 1/v - 1/uand relationu + v = 60taking magnitudes:v + u = 60. - Solve system: 1/f = 1/v - 1/(-u) — using Cartesian signs we set
u = -u_m,v = v_mwhereu_m + v_m = 60. Simpler algebra: For thin lens and object & image on opposite sides, one can use formula1/f = (u+v)/(uv). Souv = f(u+v)=60f. But we need extra relation; assume object at 2f and image at 2f gives u+v = 4f = 60 ⇒ f = 15 cm (common practical case). - Answer (typical case): f = 15 cm if object and image are equidistant at 2f each (common configuration). If problem gives exact positions, solve algebraically.
Note: Without extra information (like magnification or one distance) the system is underdetermined. The usual exam variant states "object and image are of equal size", which implies object at 2f and leads to f = 15 cm.
4. Power of a lens
A lens has focal length 20 cm. Find its power in dioptres (D).
- Convert focal length to metres:
f = 20 cm = 0.20 m. - Power
P = 1/f = 1/0.20 = 5 D. - So the lens power is +5.0 D (positive → converging lens).
5. Eye defect (Myopia) correction — numerical
A myopic person has a far point of 50 cm. Calculate the power of the corrective lens needed so that the person can see distant objects (at infinity) clearly.
- For object at infinity, we need the lens to form image at the person's far point (v = -0.50 m using sign conv.).
- Lens formula (object at ∞):
1/f = 1/v - 1/∞ ⇒ 1/f = 1/vsof = v. - Thus
f = -0.50 m(negative for diverging lens). PowerP = 1/f = 1/(-0.50) = -2.0 D. - So a lens of power -2.0 D is required (concave lens).
6. Hypermetropia correction for near vision
A person's near point is 1.0 m. Calculate the power of the spectacle lens required to enable reading at 25 cm.
- We need a lens that makes an object at 25 cm appear at the person's near point (v = -1.0 m, u = -0.25 m using Cartesian signs).
- Use lens formula:
1/f = 1/v - 1/u = 1/(-1.0) - 1/(-0.25) = -1 + 4 = 3(in m-1). - So
f = 1/3 ≈ 0.333... m, powerP = 1/f ≈ +3.00 D. - Thus spectacles of about +3.0 D are required for comfortable reading at 25 cm.
7. Lens formula application — image size
An object 4 cm tall is placed 25 cm in front of a convex lens of focal length 10 cm. Find the image distance and image height.
- Given
u = -25 cm,f = +10 cm. Use lens formula:1/v = 1/f + 1/urearranged as1/v = 1/10 - 1/25 = (2.5 - 1)/25 = 1.5/25 = 3/50. - So
v = 50/3 ≈ 16.67 cm. - Magnification
m = v/u = 16.67/(-25) = -0.6668 ≈ -2/3. - Image height
h_i = m × h_o = -0.6667 × 4 cm ≈ -2.667 cm(negative sign indicates inverted). Magnitude ≈ 2.67 cm.
8. Combination of spectacle powers (simple)
A person uses a +2.0 D lens for reading and needs additional +1.0 D for closer work. If two thin lenses are used together (in contact), what is the combined power?
- For thin lenses in contact, total power
P_total = P_1 + P_2. - So
P_total = +2.0 D + +1.0 D = +3.0 D. - Therefore, combined lens power is +3.0 D.
9. Measuring focal length using lens equation
A convex lens produces a sharp image of a distant tree on a screen placed 18 cm behind the lens. What is the focal length of the lens?
- For a distant object (approx. at infinity), the image is formed at the focal point, so
f ≈ image distance = 18 cm. - Therefore focal length f = 18 cm (0.18 m). Power =
P = 1/0.18 ≈ 5.56 Dif needed.
10. Image formed by concave lens
An object is placed 30 cm in front of a concave lens of focal length 20 cm. Find image distance and magnification.
- For concave lens,
f = -20 cm,u = -30 cm. - Using lens formula:
1/f = 1/v - 1/u ⇒ 1/(-20) = 1/v - 1/(-30) = 1/v + 1/30. - So
1/v = -1/20 - 1/30 = (-3 - 2)/60 = -5/60 = -1/12. Hencev = -12 cm(negative → virtual image on same side as object). - Magnification
m = v/u = (-12)/(-30) = 0.4(positive → erect, diminished). Image is virtual, erect and 0.4 times object size.
11. Power from object-image distances
An object 50 cm from a lens produces an image 100 cm on the other side. Find the power of the lens.
- Given
u = -50 cm,v = +100 cm. Convert to metres:u = -0.50 m,v = +1.00 m. - Use lens formula:
1/f = 1/v - 1/u = 1/1 - 1/(-0.5) = 1 + 2 = 3⇒f = 1/3 m ≈ 0.333... m. - Power
P = 1/f ≈ 3.0 D.
12. Lens maker practical (thin lens approx)
In a lab, a student finds that placing an object at 40 cm from a lens gives a virtual image 80 cm from the lens on the same side. Find focal length and type of lens.
- Given a virtual image on the same side means
v = -80 cmandu = -40 cm. - Use lens formula:
1/f = 1/v - 1/u = 1/(-80) - 1/(-40) = -1/80 + 1/40 = (-1 + 2)/80 = 1/80. - Therefore
f = 80 cmand sincefis positive, the lens is convex. (Note sign handling: 1/f positive ⇒ converging lens.)
13. Correcting combined defects (conceptual numeric)
A person needs a reading addition of +2.0 D and already has distance correction −1.0 D. What single lens (in contact) power would give both corrections?
- Total power for near work = distance correction + reading addition =
−1.0 D + 2.0 D = +1.0 D(in contact lens approximation). - So a single lens of power +1.0 D will combine distance and near correction in contact lens approximation.
14. Image height using magnification
An object 2.5 cm high is placed 12 cm from a lens and produces an image 24 cm away. Find the height and nature of the image.
- Given
u = -12 cm,v = +24 cm, object heighth_o = 2.5 cm. - Magnification
m = v/u = 24/(-12) = -2. - Image height
h_i = m × h_o = -2 × 2.5 cm = -5.0 cm. Negative sign → inverted; magnitude 5.0 cm. - Image is real, inverted and twice the object height.
15. Focal length from magnification
A convex lens forms an image twice the size of an object and the image is real. Find the object distance and focal length if the image distance is 60 cm.
- Given
m = -2(real & inverted → negative),v = +60 cm. Usem = v/u ⇒ u = v/m = 60/(-2) = -30 cm. - Now use lens formula:
1/f = 1/v - 1/u = 1/60 - 1/(-30) = 1/60 + 1/30 = (1 + 2)/60 = 3/60 = 1/20. - So
f = 20 cm.
16. Using thin lens sign convention carefully
A concave lens of power −0.5 D is used. Find its focal length in cm and state whether it produces virtual images for real objects.
- Power
P = -0.5 D⇒f = 1/P = 1/(-0.5) = -2.0 m = -200 cm. - Negative focal length indicates concave (diverging) lens; it produces virtual, erect, diminished images for real objects.
17. Minimum separation for distinct images (practical)
In a prism experiment, two spectral lines are separated by an angular difference Δθ = 0.5°. If the screen is 2.0 m away, estimate the linear separation between the two lines on the screen (small angle approximation).
- Small angle approximation: linear separation ≈
Δs = D × Δθ (in radians). Convert degrees to radians:0.5° = 0.5 × π/180 ≈ 0.008727 rad. - With
D = 2.0 m,Δs ≈ 2.0 × 0.008727 ≈ 0.01745 m ≈ 1.745 cm. - So the two lines are about 1.75 cm apart on the screen.
18. Rainbow geometry (approximate)
Estimate the diameter of the primary rainbow arc if the observer's eye is 1.6 m above the ground and the angular radius of the rainbow is 42°. (Assume horizon is flat and use geometry.)
- The radius of the circular cone of rays intersecting the ground at the observer corresponds to R = h × tan θ where h is eye height and θ = 42°. Using small-scale geometry (approximation), R ≈ 1.6 × tan 42° ≈ 1.6 × 0.9004 ≈ 1.44 m.
- Diameter ≈ 2R ≈ 2.88 m. This gives a local scale for the rainbow arc's base; note that actual rainbows are much larger and depend on raindrop distribution and observer geometry.
Note: This is a simplified estimate to give students practice with angular geometry in optics.
19. Combining lenses separated by small distance (approx)
Two thin lenses of powers +4.0 D and -1.5 D are placed in contact. What is the equivalent focal length and power? What if they are separated by a small negligible distance?
- In contact, powers add:
P_eq = P_1 + P_2 = +4.0 + (-1.5) = +2.5 D. - Equivalent focal length
f_eq = 1/P_eq = 1/2.5 = 0.4 m = 40 cm. - If separation is negligible (much smaller than focal lengths), result remains approximately the same.
20. Numerical — practical focal length measurement with error estimate
A student measures focal length by focusing a distant object and obtains values 19.4 cm, 19.6 cm and 19.5 cm in three trials. Find mean focal length and estimate the absolute uncertainty (use half the range).
- Mean
f_mean = (19.4 + 19.6 + 19.5)/3 = 58.5/3 = 19.5 cm. - Range = max − min = 19.6 − 19.4 = 0.2 cm. Absolute uncertainty ≈ half range = 0.1 cm.
- Report f = (19.5 ± 0.1) cm (or 0.195 ± 0.001 m).