Light Refraction Solved Examples (Class 10 Physics)
Master refraction calculations using Snell's law and the concept of refractive index. These problems cover light bending at interfaces and critical angle c
TL;DR: Master refraction calculations using Snell's law and the concept of refractive index. These problems cover light bending at interfaces and critical an…
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Master refraction calculations using Snell's law and the concept of refractive index. These problems cover light bending at interfaces and critical angle c
Light Refraction — Solved Numerical Examples (Step by Step)
Example 1: Light travels from air into glass. The angle of incidence is 30° and the angle of refraction is 19°. Calculate the refractive index of glass.
Solution: Using Snell's law: n1 sin(i) = n2 sin(r)
For air, n1 = 1
1 × sin(30°) = n_glass × sin(19°)
sin(30°) = 0.5
sin(19°) = 0.326
n_glass = sin(30°) / sin(19°) = 0.5 / 0.326 = 1.53
Example 2: A ray of light enters a medium from air at an angle of 45° and refracts to 30°. What is the refractive index of the medium?
Solution: Using Snell's law: n_air × sin(45°) = n_medium × sin(30°)
1 × sin(45°) = n_medium × sin(30°)
sin(45°) = 0.707, sin(30°) = 0.5
n_medium = 0.707 / 0.5 = 1.414
Example 3: Light travels from a medium of refractive index 1.5 to air. If the angle of incidence is 40°, find the angle of refraction.
Solution: Using Snell's law: n1 sin(i) = n2 sin(r)
1.5 × sin(40°) = 1 × sin(r)
1.5 × 0.643 = sin(r)
sin(r) = 0.964
r = sin⁻¹(0.964) = 74.6°
Example 4: Find the critical angle for light traveling from glass (n = 1.5) to air.
Solution: At critical angle, refracted angle = 90°
Using Snell's law: n_glass × sin(c) = n_air × sin(90°)
1.5 × sin(c) = 1 × 1
sin(c) = 1/1.5 = 0.667
c = sin⁻¹(0.667) = 41.8°
Example 5: A light ray hits a glass plate (n = 1.6) at an angle of 60° from the normal. What is the angle of refraction?
Solution: Using Snell's law: n_air × sin(60°) = n_glass × sin(r)
1 × sin(60°) = 1.6 × sin(r)
0.866 = 1.6 × sin(r)
sin(r) = 0.866 / 1.6 = 0.541
r = sin⁻¹(0.541) = 32.8°
Example 6: The refractive index of a diamond is 2.42. Calculate its critical angle.
Solution: At critical angle: n_diamond × sin(c) = n_air × sin(90°)
2.42 × sin(c) = 1 × 1
sin(c) = 1/2.42 = 0.413
c = sin⁻¹(0.413) = 24.4°
Tips
- Snell's law: n1 sin(θ1) = n2 sin(θ2) is always applicable at the interface between two media.
- Critical angle occurs only when light travels from a denser to a less dense medium; beyond this angle, total internal reflection occurs.
- Refractive index is always greater than or equal to 1 (n ≥ 1); it indicates how much a medium slows light compared to vacuum.
Frequently Asked Questions
Why does a pencil appear bent in water?
Light from the submerged part of the pencil refracts (bends) as it exits the water, making the pencil appear bent at the water surface.
What is total internal reflection?
When light traveling in a denser medium hits the interface with a less dense medium at an angle greater than the critical angle, all light is reflected back; no refraction occurs.
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