Light Exemplar — Class 10 Science NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Science chapter "Light Exemplar" — 5 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Science chapter "Light Exemplar" — 5 important questions with detailed answers for CBSE board exam prep…
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Key Questions Covered:
- A light ray hits a plane mirror at 30° to the normal. What is angle of reflec…
- An object is placed 15 cm from a concave mirror of focal length 10 cm. Find i…
- Define refractive index. If light travels 2 × 10^8 m/s in a medium, find refr…
- A light ray enters glass slab (n = 1.5) from air at 60° to normal. Find angle…
- Explain why objects in water appear closer than they actually are.
Solutions Summary:
| Question | Status |
|---|---|
| A light ray hits a plane mirror at 30° to the normal. Wha… | ✓ Solved |
| An object is placed 15 cm from a concave mirror of focal … | ✓ Solved |
| Define refractive index. If light travels 2 × 10^8 m/s in… | ✓ Solved |
| A light ray enters glass slab (n = 1.5) from air at 60° t… | ✓ Solved |
| Explain why objects in water appear closer than they actu… | ✓ Solved |
Showing 5 of 5 questions
Q1: A light ray hits a plane mirror at 30° to the normal. What is angle of reflection?
By law of reflection: angle of incidence = angle of reflection. If light hits at 30° to normal (i = 30°), then angle of reflection r = 30°.
Q2: An object is placed 15 cm from a concave mirror of focal length 10 cm. Find image position and nature.
Using mirror formula: 1/f = 1/u + 1/v. 1/10 = 1/15 + 1/v. So 1/v = 1/10 - 1/15 = (3-2)/30 = 1/30. Thus v = 30 cm. Since v is positive and real, image is real, inverted, enlarged, located at 30 cm in front of mirror.
Q3: Define refractive index. If light travels 2 × 10^8 m/s in a medium, find refractive index.
Refractive index n = speed of light in vacuum / speed of light in medium = c/v. n = (3 × 10^8)/(2 × 10^8) = 1.5. The medium is glass-like.
Q4: A light ray enters glass slab (n = 1.5) from air at 60° to normal. Find angle of refraction.
Using Snell's law: n1 sin θ1 = n2 sin θ2. 1 × sin 60° = 1.5 × sin θ2. So sin θ2 = (√3/2)/1.5 = √3/3 ≈ 0.577. Thus θ2 ≈ 35.3°.
Q5: Explain why objects in water appear closer than they actually are.
Water has refractive index 1.33 > 1 (air). When light travels from water to air (less dense to more dense), it bends away from normal. Our eyes receive refracted rays and perceive image at apparent position closer than actual position. This is due to refraction.
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