20 Numerical Problems — Topic-wise (with detailed steps)
Topic A — Shadows & Similar Triangles
1. A lamp (considered a point source) is 100 cm above the ground. A vertical pole 2 m tall stands on the ground. Find the length of the pole’s shadow on the ground when the lamp is directly above the top of the pole (lamp located above the pole top).
- Interpretation: Lamp above the pole top means rays from lamp are vertical — shadow length will be zero in ideal geometry. But if lamp is directly above the pole top and light is vertical, the pole does not cast a horizontal shadow.
Answer: 0 cm (no horizontal shadow) — because light rays fall vertically down when lamp is exactly above pole's top.
2. A small lamp is fixed 150 cm above the ground. A toy of height 30 cm is placed 50 cm from the vertical under the lamp; a screen is placed 200 cm from the lamp on the opposite side. Calculate the shadow length on the screen using similar triangles (assume lamp is point source).
- Set up similar triangles: object height / (distance from lamp to object) = shadow height on screen / (distance from lamp to screen).
- Distances from lamp: object is 50 cm below lamp horizontally? Clarify geometry: assume distances along same line: object lies on line between lamp and screen: object at 50 cm from lamp, screen at 200 cm from lamp.
- Use ratio: shadow height = object height × (distance lamp→screen) / (distance lamp→object) = 30 × (200/50) = 30 × 4 = 120 cm.
Answer: 120 cm shadow height on the screen (the projected image appears larger due to geometry).
3. A torch acts like a point source located 40 cm from a small object of height 8 cm. A screen is placed 120 cm from the torch on the other side. Find the height of the object’s shadow on the screen.
- Use similar triangles: object height / distance (torch→object) = shadow height / distance (torch→screen).
- Compute: shadow = 8 × (120/40) = 8 × 3 = 24 cm.
Answer: 24 cm.
4. A student places a meter stick vertically 80 cm from a point light source. A screen is placed 200 cm from the same light source. What is the length of the shadow?
- Similar triangles: shadow = object height × (screen distance / object distance) = 100 cm × (200/80) = 100 × 2.5 = 250 cm.
Answer: 250 cm (2.5 m).
5. A lamp of finite size (an extended source of vertical length 6 cm) is located 120 cm from a small object. Using a simple model, estimate qualitatively whether the shadow will have a noticeable penumbra and why.
- Because the source has finite size, different parts of the source will be blocked/unblocked producing partial shading.
- If the source size compared to distances is not negligible, penumbra is noticeable.
Answer: Yes — the shadow will have a noticeable penumbra because the lamp is extended (6 cm) and at a moderate distance, causing partial illumination at edges.
Topic B — Reflection & Angles
6. A ray of light strikes a plane mirror at an angle of 35° to the normal. Find the angle the reflected ray makes with the mirror surface.
- Angle of reflection = angle of incidence = 35° (measured from normal).
- Angle with mirror surface = 90° − angle with normal = 90° − 35° = 55°.
Answer: The reflected ray makes 55° with the mirror surface (and 35° with the normal).
7. A student traces an incident ray hitting a mirror. If the incident ray is at 20° to the mirror surface, what is the angle of reflection measured from the surface?
- Convert to angle with normal: angle with normal = 90° − 20° = 70°.
- Angle of reflection with normal = 70°; convert back to surface: angle with surface = 90° − 70° = 20°.
Answer: The reflected ray is also at 20° to the mirror surface (symmetry: angle of incidence = reflection).
8. A ray strikes a plane mirror. The reflected ray is observed to be 40° above the mirror surface. What was the angle of incidence with the normal?
- Angle of reflection with surface = 40°. Angle with normal = 90° − 40° = 50°.
- Angle of incidence = angle of reflection = 50° (with normal).
Answer: 50° with the normal.
9. Two plane mirrors are placed at right angles (90°). A ray strikes the first mirror, reflects, then reflects off the second mirror. If the incident ray strikes the first mirror at 30° to the normal, what is the direction of the final emerging ray relative to the original (qualitative answer)?
- Reflection from first mirror reverses angle; then reflection from second mirror causes another symmetric change. For two perpendicular mirrors, the emerging ray is parallel to the incident ray but displaced (essential property of two reflections at 90°).
Answer: The final ray emerges parallel to the original incident ray (but displaced), a key property used in periscopes and corner reflectors.
10. A periscope uses mirrors at 45°. A horizontal ray enters the top opening and reflects down the tube; calculate the total change in direction (degrees) after two reflections.
- A 45° mirror redirects horizontal ray down by 90° after first reflection. The second mirror redirects the downward ray to horizontal again but towards the eye — net direction change is 0° relative to initial horizontal direction but path is inverted vertically; total turns are two 90° bends summing 180° of turning.
Answer: The ray is turned down 90° by the first mirror and back to horizontal by the second (total bending actions 90° + 90° = 180°); overall the outgoing ray is parallel to the incoming ray.
Topic C — Plane Mirrors & Image Geometry
11. An object 30 cm tall is placed 60 cm in front of a plane mirror. Where will the image be, and what will be its size?
- Plane mirror property: image distance = object distance → image will be 60 cm behind the mirror.
- Image size = object size for plane mirror → 30 cm tall.
Answer: Image is 60 cm behind the mirror and 30 cm tall (virtual and erect).
12. A boy 1.6 m tall stands 2 m in front of a plane mirror. How far behind the mirror does his image appear? How tall is the image?
- Image distance = object distance = 2 m behind the mirror.
- Image height = object height = 1.6 m.
Answer: Image 2 m behind the mirror and 1.6 m tall (virtual, erect).
13. A mirror on a wall is 1.2 m tall. What is the maximum height of a person who can see their full body in this mirror when standing at an appropriate distance? (Hint: A mirror of height half the person's height is enough to see full body if positioned properly.)
- Rule: A vertical plane mirror of half the height of the person is sufficient to see full-length image (when placed appropriately).
- If mirror height = 1.2 m, maximum person height = 2 × 1.2 = 2.4 m.
Answer: Up to 2.4 m tall person can see full body in a 1.2 m tall mirror (practical placements assumed).
14. An object is 50 cm in front of a plane mirror. A camera placed 10 cm behind the mirror records the virtual image. What is the distance between the camera and the real object (straight-line), if camera, mirror and object are all on same axis?
- Object is 50 cm in front of mirror → image is 50 cm behind mirror.
- Camera is 10 cm behind mirror → distance from camera to image = 50 − 10 = 40 cm (but camera records image; distance camera to object along axis = camera to mirror + mirror to object = 10 + 50 = 60 cm).
Answer: Camera is 60 cm from the real object along the axis (10 cm behind mirror + 50 cm in front of mirror).
Topic D — Pinhole Projection & Scaled Images
15. In a pinhole setup, an object 20 cm tall is placed 100 cm from the pinhole. The screen is 10 cm on the other side of the pinhole. Estimate the height of the image on the screen (use similar triangles).
- Similar triangles: image height / object height = screen distance from pinhole / object distance from pinhole = 10/100 = 1/10.
- Image height = 20 × (1/10) = 2 cm.
Answer: 2 cm tall image (inverted).
16. A card with a 0.5 cm pinhole forms a 3 cm tall inverted image of a distant object on a screen 5 cm behind the pinhole. Estimate how far the object is from the pinhole (assume object height 150 cm).
- Similar triangles: image height / object height = screen distance / object distance ⇒ 3/150 = 5 / D ⇒ D = 5 × (150/3) = 5 × 50 = 250 cm.
Answer: Object approximately 250 cm from the pinhole.
Topic E — Practical Measurements & Estimations
17. A 1.5 m tall flagpole casts a 3 m long shadow at a certain time. Later, the shadow length reduces to 1.5 m. Assuming sun rays are parallel, how did the sun’s elevation angle change qualitatively between the two times?
- Shadow length is inversely related to sun elevation: longer shadow means lower sun angle; shorter shadow means higher sun angle.
Answer: The sun’s elevation angle increased between the two times (shadow shortened from 3 m to 1.5 m, so sun moved higher).
18. A student measures the shadow of a 2 m tall tree to be 8 m. What is the angle of elevation of the sun (use tan θ = opposite/adjacent with small-angle approximation)?
- Here, tan θ = height / shadow length = 2/8 = 0.25 ⇒ θ = arctan(0.25). Approx θ ≈ 14.04° (using table or calculator).
Answer: θ ≈ 14.0° (angle of elevation).
19. A torch placed at floor level projects light onto a wall 3 m away. A small toy 30 cm tall is 1 m from the torch. Determine the height of the shadow on the wall using similar triangles.
- Distances from torch: toy at 1 m, wall at 3 m ⇒ ratio = 3/1 = 3.
- Shadow height = object height × ratio = 30 cm × 3 = 90 cm.
Answer: 90 cm tall shadow on the wall.
20. In a classroom demonstration, a student places two identical lamps on opposite sides of an object producing two equal overlapping shadows. If one lamp is moved farther away by doubling its distance, describe qualitatively how its shadow contribution changes.
- When distance increases, the shadow produced by that lamp becomes smaller and less intense on the screen (for point-source model) because the angular size of the object as seen from the lamp decreases.
Answer: The lamp moved farther will cast a smaller and fainter shadow; the nearer lamp will dominate the shadow pattern.
Teacher’s tip: Encourage students to sketch the geometry for each numerical problem — draw rays, mark distances and use similar triangles. For mirror-angle problems, always draw the normal and measure angles from it.