Unit 14: Light

Long Questions & Detailed Answers

Based on National Curriculum 2023 | PECTAA 2026 Syllabus

✍️ Prepared by Muhammad Tayyab

🏫 Subject Specialist Physics | Govt Christian High School Daska

πŸ’‘ Chapter 14: Light – Long Questions

Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).

πŸ“– What's Inside: This chapter covers reflection, critical angle, total internal reflection, optical fibres, long-sightedness and their applications. Each long question is presented with the exact exam-ready answer as per the official PECTAA 2026 Physics curriculum.

⬇️ Download PDF (Long Questions)

πŸ“š Related Resources – Chapter 14: Light

Light covers reflection, refraction, total internal reflection, optical fibres, lenses, prisms, dispersion, and optical instruments.

πŸ“‘ Quick Jump to Questions

πŸ“– Long Questions & Answers (PECTAA 2026)

14.1 Explain the terms; normal, angle of incidence and angle of reflection in the context of light reflection. Provide a diagram to illustrate these terms.

Normal: A line drawn perpendicular (90Β°) to the reflecting surface at the point where the incident ray strikes it.

Angle of Incidence (i): The angle between the incident ray and the normal at the point of incidence.

Angle of Reflection (r): The angle between the reflected ray and the normal at the point of reflection.

Law of Reflection: The incident ray, the normal, and the reflected ray all lie in the same plane, and the angle of incidence is equal to the angle of reflection (i = r).
14.2 Define the terms critical angle and total internal reflection. Explain the conditions required for total internal reflection to occur.

Critical Angle: The angle of incidence for which the angle of refraction is 90Β° is called the critical angle.

Total Internal Reflection (TIR): Total internal reflection is the phenomenon of reflection of light ray back to the same medium when passing from denser medium to rarer medium in such a way that angle of incidence is greater than its critical angle.

Conditions Required for Total Internal Reflection:
i. Light must travel from a denser medium to a rarer medium (higher refractive index to lower refractive index).
ii. The angle of incidence must be greater than the critical angle (i > C).
14.3 Derive the equation; \(n = \frac{1}{\sin C}\) for the refractive index in terms of the critical angle C. Include all steps and assumptions.

Assumptions:
i. Light travels from a denser medium (refractive index n1 = n) to air (refractive index n2 = 1).
ii. The angle of incidence is exactly equal to the critical angle (i = C).

Derivation: To derive a formula for total internal reflection, we start with Snell's law:

\[ n_1 \sin i = n_2 \sin r \]

Substitute the given conditions into the equation:
β€’ Let n1 = n (refractive index of the denser medium)
β€’ Let n2 = 1 (refractive index of air)
β€’ Set the angle of incidence i = C (critical angle)
β€’ By definition of the critical angle, the angle of refraction r = 90Β°

\[ n \sin C = 1 \sin 90Β° \]
\[ n \sin C = (1)(1) \]
\[ n \sin C = 1 \]
\[ n = \frac{1}{\sin C} \]

Finding the Critical Angle: if we have to find critical angle, then

\[ \sin C = \frac{1}{n} \]
\[ C = \sin^{-1} \left( \frac{1}{n} \right) \]

Note: These relations are valid when light travels from a denser medium to air (or vacuum). More generally, if light travels from a medium of refractive index n1 to another medium of refractive index n2, then

\[ C = \sin^{-1} \left( \frac{n_2}{n_1} \right) \]
14.4 Describe the structure of an optical fibre. Write its any two advantages.

An optical fibre is a very thin, flexible strand made of high-quality glass or plastic that transmits light signals.

It consists of two main parts:

Core: The central inner cylinder made of glass or plastic with a higher refractive index. Light signals travel inside this core.

Cladding: An outer layer surrounding the core, made of glass or plastic with a lower refractive index.

This difference in refractive indices allows total internal reflection to keep the light contained inside the core.

Two Advantages of Optical Fibres: Compared to traditional copper cables, optical fibres offer several key benefits:

i. High Bandwidth: Optical fibres can carry a much larger amount of data than copper wires. Bandwidth means how much data can be sent over a network per second. In today's world, where we need fast internet, optical fibres are the best choice because they support high-speed data transfer.

ii. Low Power Consumption: They use less power than copper cables. Also, because they last longer and are more durable, they reduce the cost of repairs and maintenance.

iii. Faster Speed: They send data using light pulses, which travel extremely fast almost at the speed of light.

iv. Long Distance Transmission: Optical fibres can carry data across very long distances without losing signal quality.

v. Resistance to Electrical Interference: Since optical fibres use light instead of electricity, they are not affected by electrical noise or interference.
14.5 What is long-sightedness? Discuss its causes.

In long-sightedness, a person can see distant objects clearly, but nearby objects appear blurry.

Causes of Long-sightedness: This happens when the eye lens is too weak or the eyeball is too small, so the light rays from near objects are not bent enough. As a result, the rays focus behind the retina instead of on it.

There are two main causes:

i. Lens too weak: The lens does not bend the light rays enough.

ii. Eyeball too small: The retina is too close to the lens.

In both cases, the image is formed behind the retina, making close-up vision blurry.

πŸ“ Key Formulas – Light

Refractive Index: \( n = \frac{c}{v} \)
Snell's Law: \( n_1 \sin i = n_2 \sin r \)
Critical Angle: \( C = \sin^{-1} \left( \frac{1}{n} \right) \)
Mirror / Lens Formula: \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \)
Magnification: \( M = \frac{h_i}{h_o} = -\frac{q}{p} \)
Power of Lens: \( P = \frac{1}{f} \)

πŸ“– Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21

πŸ’‘ Exam Tip:

For board exams, define key terms precisely, mention formulas with units, and relate to real-life examples. These long questions follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.

Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus

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