Skip to content
Login Register

#wave optics

10 public questions tagged with this topic.

Total Internal Reflection and Critical Angle

What is the primary reason optical fibers are bent without losing light transmission efficiency?

**Critical angle** C satisfies sinC = n₂/n₁, n₁>n₂, n₂=1 for air, n₁=1.52 for glass gives sinC=1/1.52=0.6579, C≈41.1°, for water n=1.33 C=48.75°. Beyond C, total internal reflection occurs, all light reflected, no refracted ray, used in optical fibers and prisms. Optical fibers maintain light transmission when bent because the angle of incidence at the core-cladding interface remains greater than the critical angle, ensuring total internal reflection. This allows light to follow the bend without escaping into the cladding. Substituting values gives Due to sustained total internal reflection, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f =

Ref: NCERT > Physics Book > Ray Optics > Total Internal Reflection and Critical Angle

View Details
PHYSICS

What is the phase difference corresponding to a path difference of 7lambda/4 in a double-slit experiment?

Given: What is the phase difference corresponding to a path difference of 7lambda/4 in a double-slit experiment? These values define the system as per NCERT data. Formula: Phase difference phi = 2π/lambda Δ. This is standard NCERT relation. Substitution & Calculation: For Δ = 7lambda/4, phi = 2π/lambda · 7lambda/4 = 7π/2 . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Wave Optics, Topic: Interference, phase difference φ = (2π/λ)Δ, path difference 7λ/4 and double-slit experiment. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

View Details
PHYSICS

In a double-slit experiment, if lambda = 450 nm, d = 0.15 mm, and D = 1.5 m, what is the fringe width?

Given: In a double-slit experiment, if lambda = 450 nm, d = 0.15 mm, and D = 1.5 m, what is the fringe width? These values define the system as per NCERT data. Formula: Fringe width β = lambda D/d. This is standard NCERT relation. Substitution & Calculation: lambda = 4.5 × 10⁻⁷m, d = 1.5 × 10⁻⁴m, D = 1.5 m . β = frac4.5 × 10⁻⁷ × 1.51.5 × 10⁻⁴= 4.5 × 10⁻³m = 4.5 mm . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Wave Optics, Topic: Interference, phase difference φ = (2π/λ)Δ, path difference 7λ/4 and double-slit experiment. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

View Details
PHYSICS

What is the angular position of the first minimum in a single-slit diffraction pattern if the slit width is 10.0 μm and

Given: What is the angular position of the first minimum in a single-slit diffraction pattern if the slit width is 10.0 μm and the wavelength is 500 nm ? These values define the system as per NCERT data. Formula: First minimum occurs at sin θ = lambda/a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 5.0 × 10⁻⁷ m, a = 1.0 × 10⁻⁵ m . sin θ = frac5.0 × 10⁻⁷¹.0 × 10⁻⁵= 0.05, θ = sin^{-1(0.05) approx 2.9° . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Ray Optics and Optical Instruments and Wave Optics, Topic: Refraction, lenses and interference/diffraction.

View Details
PHYSICS

In a double-slit experiment, if lambda = 460 nm, d = 0.2 mm, and D = 2.0 m, what is the distance of the third bright fri

Given: In a double-slit experiment, if lambda = 460 nm, d = 0.2 mm, and D = 2.0 m, what is the distance of the third bright fringe from the ntral maximum? These values define the system as per NCERT data. Formula: Bright fringe position x_n = n lambda D/d. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For the third bright fringe, n = 3 . lambda = 4.6 × 10⁻⁷ m, d = 2.0 × 10⁻⁴ m, D = 2.0 m . x_3 = frac3 × 4.6 × 10⁻⁷ × 2.02.0 × 10⁻⁴= 6.9 × 10⁻³ m = 6.9 mm . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

View Details
PHYSICS

What ensures that the laws of reflection hold true when derived using the wave model?

The wave model uses secondary wavelets, where the reflected wavefront’s envelope forms when the incident and reflected angles are equal, satisfying geometric congruence.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Wave Optics, Topic: Laws of reflection from wave model, secondary wavelets and equal angles. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

View Details
PHYSICS

What is the phase difference corresponding to a path difference of 5lambda/4 in a double-slit experiment?

Given: What is the phase difference corresponding to a path difference of 5lambda/4 in a double-slit experiment? Formula: Phase difference phi = 2π/lambda Δ. Substitution & Calculation: For Δ = 5lambda/4, phi = 2π/lambda · 5lambda/4 = 5π/2 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27), Chapter: Wave Optics (Latest NCERT 2026-27), Topic: Interference, phase difference φ = (2π/λ)Δ, path difference 5λ/4 and double-slit experiment. The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI units and illustrative examples. Page number.

View Details