Light Reflection and Refraction Solved Examples (Class 10 Physics)
Light travels in straight lines and can be reflected or refracted when it hits surfaces or boundaries. The laws of reflection and Snell's law help us predi
TL;DR: Light travels in straight lines and can be reflected or refracted when it hits surfaces or boundaries. The laws of reflection and Snell's law help us…
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Light travels in straight lines and can be reflected or refracted when it hits surfaces or boundaries. The laws of reflection and Snell's law help us predi
Light Reflection and Refraction — Solved Numerical Examples (Step by Step)
Example 1: A light ray hits a plane mirror at an angle of incidence of 30 degrees. What is the angle of reflection?
Solution: Given:
Angle of incidence i = 30 degrees
According to the law of reflection:
Angle of incidence = Angle of reflection
Angle of reflection = 30 degrees
Example 2: Light travels from air into water with an angle of incidence of 45 degrees. The refractive index of water is 1.33. Calculate the angle of refraction.
Solution: Given:
Angle of incidence i = 45 degrees
Refractive index of air n1 = 1
Refractive index of water n2 = 1.33
Using Snell's law: n1 sin i = n2 sin r
1 × sin 45 = 1.33 × sin r
0.707 = 1.33 × sin r
sin r = 0.707 / 1.33
sin r = 0.532
r = 32.1 degrees (approximately)
Example 3: An object is placed 15 cm in front of a concave mirror with a focal length of 10 cm. Calculate the image distance.
Solution: Given:
Object distance u = 15 cm
Focal length f = 10 cm
Using the mirror formula: 1/f = 1/u + 1/v
1/10 = 1/15 + 1/v
1/v = 1/10 - 1/15
1/v = 3/30 - 2/30
1/v = 1/30
v = 30 cm
Example 4: Calculate the magnification when an object 5 cm tall is placed 20 cm from a concave mirror with focal length 8 cm.
Solution: Given:
Object height h = 5 cm
Object distance u = 20 cm
Focal length f = 8 cm
First find image distance using 1/f = 1/u + 1/v:
1/8 = 1/20 + 1/v
1/v = 1/8 - 1/20 = 5/40 - 2/40 = 3/40
v = 40/3 = 13.33 cm
Magnification m = -v/u = -13.33/20 = -0.667
Example 5: A convex lens has a focal length of 20 cm. An object is placed 30 cm from the lens. Find the image distance.
Solution: Given:
Focal length f = 20 cm
Object distance u = 30 cm
Using lens formula: 1/f = 1/u + 1/v
1/20 = 1/30 + 1/v
1/v = 1/20 - 1/30
1/v = 3/60 - 2/60
1/v = 1/60
v = 60 cm
Example 6: Light travels from a denser medium with refractive index 1.5 to air with refractive index 1. If the angle of incidence is 30 degrees, find the angle of refraction.
Solution: Given:
Refractive index of denser medium n1 = 1.5
Refractive index of air n2 = 1
Angle of incidence i = 30 degrees
Using Snell's law: n1 sin i = n2 sin r
1.5 × sin 30 = 1 × sin r
1.5 × 0.5 = sin r
sin r = 0.75
r = 48.6 degrees (approximately)
Example 7: A plane mirror is tilted at 10 degrees from its original position. By what angle will the reflected ray turn?
Solution: Given:
Tilt angle of mirror = 10 degrees
When a mirror is tilted by an angle theta, the reflected ray turns by 2 × theta
Angle turned = 2 × 10 = 20 degrees
Tips
- Remember the law of reflection: angle of incidence equals angle of reflection, and both are measured from the normal to the surface.
- For Snell's law, always ensure you use sine of the angles and maintain consistency with which medium has which refractive index.
- The focal length of a concave mirror is positive, and for a convex mirror it is negative. This affects the sign in the mirror formula.
Frequently Asked Questions
What is critical angle and when does total internal reflection occur?
The critical angle is the angle of incidence at which light traveling from a denser to a rarer medium refracts at 90 degrees. Beyond this angle, total internal reflection occurs instead of refraction. It is found using sin(c) = n2/n1 where n1 is denser and n2 is rarer.
How do I know if an image is real or virtual?
For mirrors and lenses, if the image distance v is positive, the image is real and inverted. If v is negative, the image is virtual and upright. Real images can be projected on a screen, virtual images cannot.
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