Reflection — Set 4

Correct Answer: A. Rough and irregular

• **Rough and irregular** = A rough surface has many tiny facets with different orientations; each facet reflects light at its own local normal, scattering the reflected rays in all directions. • **Laws obeyed locally** — diffuse reflection still obeys the laws of reflection at each infinitesimal point; it is the variation in surface normals that causes the scattering. • 💡 Wrong-option analysis: Transparent: a transparent surface transmits most light rather than reflecting it; Highly polished: a polished surface produces specular (regular) reflection, not diffuse; Perfectly smooth: a perfectly smooth surface produces ideal specular reflection with a clear image, the opposite of diffuse reflection.

Correct Answer: B. 16.5 m

• **16.5 m** = Minimum echo distance: d = v × t_min / 2 = 330 × 0.1 / 2 = 16.5 m. • **Half of total path** — sound must travel to the wall and back (total = 2d) in 0.1 s; dividing by 2 gives the one-way distance of 16.5 m. • 💡 Wrong-option analysis: 33 m: this is v × t_min = 330 × 0.1 without dividing by 2 for the return trip; 3.3 m: this is v × t_min / 10, an arithmetic error; 165 m: this is v × t_min × 5, an over-estimate with no physical basis.

Correct Answer: B. Perpendicular to the mirror surface

• **Perpendicular to the mirror surface** = The normal at any point on a reflecting surface is defined as the line perpendicular (at 90°) to the surface at that point. • **Reference for angles** — both the angle of incidence and the angle of reflection are measured from this normal, not from the surface itself. • 💡 Wrong-option analysis: At 45° to the surface: 45° would be an oblique line, not the perpendicular used to define reflection angles; Always along the principal axis: the principal axis is a concept for spherical mirrors only; for a plane mirror there is no single principal axis; Parallel to the mirror surface: a line parallel to the surface is the surface direction itself, not the normal — these are perpendicular to each other.

Correct Answer: A. Back along the same path

• **Back along the same path** = A ray along the normal has angle of incidence = 0°; by the first law of reflection angle of reflection = 0°, so the ray reflects straight back. • **Retroreflection** — this is the basis of corner reflectors and laser retroreflectors; a ray hitting any mirror at normal incidence always returns on its incident path. • 💡 Wrong-option analysis: At 90° to the incident ray: 90° between rays requires an incidence angle of 45°, not 0°; In a random direction: reflection always obeys the laws regardless of incidence angle; it never scatters randomly for a smooth mirror; Only if the mirror is concave: the first law of reflection applies to all smooth mirrors — plane, concave, and convex — not only concave mirrors.

Correct Answer: B. It follows the laws at each point but scatters due to varying normals

• **It follows the laws at each point but scatters due to varying normals** = At each tiny patch of a rough surface, the local angle of incidence equals the local angle of reflection; the overall scattering occurs because the normals point in many different directions. • **Lambertian scattering** — in ideal diffuse reflection, the scattered light is distributed uniformly over a hemisphere; this is why rough objects are visible from all angles without glare. • 💡 Wrong-option analysis: It does not follow the laws of reflection: the laws of reflection hold at every point on any surface; diffuse reflection is a macroscopic effect of microscopic roughness; It happens only in liquids: diffuse reflection occurs from any rough solid or liquid surface; It always produces a sharp image: diffuse reflection scatters rays in many directions and cannot form a sharp image — that requires specular reflection.

Correct Answer: B. Reflection of ultrasonic waves and echo timing

• **Reflection of ultrasonic waves and echo timing** = SONAR (Sound Navigation And Ranging) emits ultrasonic pulses into water; the time for the echo to return is used to calculate the distance to submerged objects. • **d = v × t / 2** — the same echo formula applies; the speed of sound in water (~1500 m/s) is used to convert the round-trip time into distance. • 💡 Wrong-option analysis: Refraction of visible light: SONAR uses sound waves, not light; refraction of light is used in optical instruments, not SONAR; Dispersion of sound in air only: SONAR operates in water, not air, and uses reflection, not dispersion; Diffraction of radio waves: radio-wave diffraction is the principle behind RADAR, not SONAR.

Correct Answer: A. -2

• **−2** = Mirror formula: 1/v = 1/f − 1/u = 1/(−10) − 1/(−15) = −1/10 + 1/15 = −1/30, so v = −30 cm; m = −v/u = −(−30)/(−15) = −2. • **m = −2** — the negative sign confirms the image is real and inverted; the magnitude 2 means it is twice the size of the object. • 💡 Wrong-option analysis: −0.5: a magnification of −0.5 would mean a diminished real image, which requires the object beyond 2f; here u = 15 cm = 1.5f, between f and 2f; +0.5: positive magnification means erect virtual image; an object at 15 cm in front of a concave mirror of f = 10 cm forms a real image; +2: positive 2 would mean virtual and erect, but the object is beyond the focus so the image is real and inverted.

Correct Answer: D. Greater intensity

• **Greater intensity** = Hard, rigid walls have a high acoustic impedance mismatch with air, causing most sound energy to be reflected rather than absorbed. • **Absorption coefficient** — soft curtains have a high acoustic absorption coefficient (near 1), converting sound to heat; hard plaster walls have a low coefficient (near 0.02), reflecting most sound. • 💡 Wrong-option analysis: No reflection at all: all surfaces reflect some sound; only a perfectly anechoic surface would have zero reflection; Only higher frequency reflection: hard walls reflect sound across a broad frequency range, not only higher frequencies; Less intensity: it is the soft curtain that reflects with less intensity (absorbs more), not the hard wall.

Correct Answer: C. 4 cm toward the object

• **4 cm toward the object** = The image is always behind the mirror at the same distance as the object is in front; when the mirror moves 2 cm toward the object, the object distance decreases by 2 cm and the image distance behind decreases by 2 cm, so the image moves 4 cm toward the object. • **Image shift = 2 × mirror shift** — because both the object-to-mirror and mirror-to-image distances change by the same amount, the image shifts twice as far as the mirror. • 💡 Wrong-option analysis: 2 cm away from the object: the mirror moving toward the object brings the image closer, not farther; 3 cm toward the object: there is no physical reason for a 3 cm shift; the factor must be 2; 2 cm toward the object: this equals the mirror's shift and ignores the doubling effect.

Correct Answer: D. 60°

• **60°** = The angle between incident and reflected rays equals twice the angle of incidence: 2i = 120°, so i = 60°. • **i = r = 60°** — both rays make 60° with the normal; the normal bisects the 120° angle between them, so each subtends 60° on either side. • 💡 Wrong-option analysis: 30°: if i = 30° the angle between rays = 2 × 30° = 60°, not 120°; 120°: 120° is the angle between the two rays, not the angle of incidence itself; 45°: if i = 45° the angle between rays = 90°, not 120°.