Light - Previous Year Questions Solved Examples (Class 10 Science)
Light reflection and refraction are fundamental concepts tested across CBSE board exams. These questions cover ray diagrams, lens formulas, and real-world
TL;DR: Light reflection and refraction are fundamental concepts tested across CBSE board exams. These questions cover ray diagrams, lens formulas, and real-w…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
Light reflection and refraction are fundamental concepts tested across CBSE board exams. These questions cover ray diagrams, lens formulas, and real-world
Light - Previous Year Questions — Solved Numerical Examples (Step by Step)
Example 1: A concave mirror forms a real, inverted, and magnified image of an object placed at a distance of 20 cm from the mirror. Where should the object be placed so that the image formed is real, inverted, and of the same size as the object? [3 marks]
Solution: For a real, inverted, magnified image, the object must be between the focus and centre of curvature (20 cm < distance < 40 cm for this mirror). For an image of the same size as the object, the object must be placed at the centre of curvature. Using the mirror formula: 1/f + 1/u = 1/v. When u = v (same size image), the object distance equals the image distance. Since the object is 20 cm away and forms a magnified image, we need it at the centre of curvature. For magnification of 1 (same size), object distance = 2f = 40 cm.
Example 2: A ray of light travels from a denser medium to a rarer medium. Will the ray bend towards the normal or away from the normal? Draw a diagram. [2 marks]
Solution: When light travels from a denser medium (higher refractive index) to a rarer medium (lower refractive index), the speed of light increases. According to Snell's law: n1 sin(theta1) = n2 sin(theta2). Since n1 > n2, we have sin(theta2) > sin(theta1), which means theta2 > theta1. The refracted ray bends away from the normal. [Diagram would show incident ray, normal, refracted ray bending away].
Example 3: An object is placed 30 cm in front of a convex lens of focal length 10 cm. Calculate the position and nature of the image formed. [3 marks]
Solution: Using lens formula: 1/f = 1/v - 1/u. Given: f = +10 cm (convex), u = -30 cm (object distance, negative by sign convention). 1/10 = 1/v - 1/(-30). 1/10 = 1/v + 1/30. 1/v = 1/10 - 1/30 = 3/30 - 1/30 = 2/30 = 1/15. Therefore v = +15 cm. Magnification m = v/u = 15/(-30) = -0.5. The image is real, inverted, and diminished.
Example 4: Why does a glass prism disperse white light into its constituent colours? [2 marks]
Solution: Different colours of white light have different wavelengths. The refractive index of glass is different for different wavelengths. Violet light (shorter wavelength) has a higher refractive index than red light (longer wavelength). Therefore, violet light bends more than red light when passing through the prism, causing the light to disperse into a spectrum.
Example 5: State Snell's law and explain its application in optical fibres. [3 marks]
Solution: Snell's law: n1 sin(i) = n2 sin(r), where n1 and n2 are refractive indices and i and r are angles of incidence and refraction. In optical fibres, light travels through a core of high refractive index surrounded by cladding of lower refractive index. When light hits the core-cladding boundary at an angle greater than the critical angle, total internal reflection occurs, confining the light within the fibre. This allows light signals to travel long distances with minimal loss.
Example 6: A concave lens always forms a virtual image. Explain why. [2 marks]
Solution: A concave lens has a negative focal length. The light rays diverge after passing through it, appearing to come from a point on the same side as the object. Using the lens formula 1/f = 1/v - 1/u, with f negative and u negative, v will always be negative, indicating the image forms on the same side as the object. A virtual image cannot be projected on a screen as the rays don't actually converge.
Example 7: An object 4 cm tall is placed 25 cm from a concave mirror of focal length 15 cm. Find the size and nature of the image. [3 marks]
Solution: Using mirror formula: 1/f + 1/u = 1/v. 1/15 + 1/(-25) = 1/v. 1/15 - 1/25 = 1/v. (5-3)/(75) = 1/v. 2/75 = 1/v. v = 37.5 cm. Magnification m = -v/u = -37.5/(-25) = 1.5. Height of image = 4 × 1.5 = 6 cm. Image is real, inverted, and magnified.
Tips
- Always use the sign convention: object distance negative, focal length negative for mirrors/concave lenses, positive for convex lenses.
- Magnification m = -v/u (mirrors) and m = v/u (lenses) helps determine if image is real or virtual.
- Draw ray diagrams carefully with at least two rays to verify the position and nature of image.
- Critical angle problems require understanding of total internal reflection condition: sin(c) = n2/n1.
Frequently Asked Questions
When does a concave mirror form a virtual image?
When the object is placed between the pole and focus (u < f), a virtual, erect, and magnified image forms behind the mirror.
What is the difference between real and virtual images?
Real images form where light rays actually converge and can be projected on a screen. Virtual images form where rays appear to come from and cannot be projected.
More Science Solved Examples
- Electricity - Previous Year Questions
- Life Processes - Previous Year Questions
- Chemical Reactions and Equations - Previous Year Questions
- Acids Bases and Salts - Previous Year Questions
- Matter in Our Surroundings - Previous Year Questions
- Force and Pressure
🤖 Stuck on any of these? Ask Syllab's free AI Tutor to explain step by step →