Questions & explanations
1. A concave mirror of focal length 20 cm and a convex mirror of focal length 30 cm are placed coaxially 50 cm apart. An object is placed 30 cm in front of the concave mirror. What is the nature and position of the final image relative to the convex mirror?
- Real, 10 cm in front of the convex mirror
- Virtual, 15 cm behind the convex mirror
- Real, 15 cm in front of the convex mirror
- Virtual, 10 cm behind the convex mirror
Answer: Real, 15 cm in front of the convex mirror
Using mirror equation for concave mirror (f=-20 cm, u=-30 cm): 1/v1 = 1/f - 1/u = -1/20 + 1/30 = -1/60, so v1 = -60 cm (real image 60 cm in front of concave mirror). This image is 10 cm behind convex mirror (since separation 50 cm), so u2 = +10 cm (virtual object). For convex mirror (f=+30 cm): 1/v2 = 1/f - 1/u2 = 1/30 - 1/10 = -1/15, so v2 = -15 cm. Negative v2 indicates real image, 15 cm in front of convex mirror.
2. A fish is at depth 2 m in water (n = 1.33). A coin lies at the bottom directly below at depth 3 m. Find the minimum radius of an opaque disc placed on the water surface to hide the coin from view.
- 2.28 m
- 3.42 m
- 1.14 m
- 4.56 m
Answer: 3.42 m
The disc must block all rays from the coin that could reach the fish. The limiting ray from the coin refracts at the critical angle at the surface. For the coin at depth 3 m, the radius needed is R = 3 * tan(i_c) where sin(i_c)=1/1.33, so i_c≈48.8°, tan≈1.14, giving R≈3.42 m. The fish's depth is irrelevant because the disc is on the surface and must cover the entire area from which rays from the coin can emerge.
3. A ray enters a 45-90-45 glass prism perpendicular to the hypotenuse. How many total internal reflections occur inside the prism?
- 1
- 2
- 3
- 0
Answer: 1
The ray enters perpendicular to the hypotenuse, so it goes straight to the opposite corner. It strikes the first short face at 45°, which is greater than the critical angle for glass (≈42°), so it undergoes total internal reflection. The reflected ray then goes to the other short face, but now the angle of incidence is 0° (normal incidence), so it exits without any further TIR. Hence only one TIR occurs.
4. An object is placed 10 cm behind a glass slab of thickness 4 cm and refractive index 1.5. The slab is followed by a spherical refracting surface of radius 20 cm separating air (n=1) and glass (n=1.5). The slab's right face coincides with the pole of the spherical surface. Where is the final image formed?
- 12.0 cm to the left of the spherical surface
- 16.6 cm to the right of the spherical surface
- 20.0 cm to the left of the spherical surface
- 16.6 cm to the left of the spherical surface
Answer: 16.6 cm to the left of the spherical surface
The slab shifts the object by t(1-1/n)=4(1-1/1.5)=4/3≈1.333 cm toward the observer, so effective object distance from the pole is u = -(10-1.333) = -8.667 cm. Using single-surface formula n2/v - n1/u = (n2-n1)/R with n1=1, n2=1.5, R=+20 cm gives 1.5/v + 1/8.667 = 0.5/20 → 1.5/v = 0.025 - 0.1154 = -0.0904 → v = -16.6 cm, meaning image is 16.6 cm left of the surface.
5. In a cold region, a distant ship appears suspended above the horizon due to looming. This happens because light from the ship undergoes total internal reflection in air layers where the refractive index:
- is highest at the surface and lowest at the top
- increases with height
- is constant with height
- decreases with height
Answer: decreases with height
In cold regions, the air near the surface is cold and dense (high refractive index), while the air above is warmer and less dense (low refractive index). So refractive index decreases with height. Light from the ship travels upward from denser to rarer medium, bending away from the normal, and eventually TIR occurs, making the ship appear above its actual position.
6. A desert mirage occurs when light from the sky undergoes total internal reflection in layers of air. This happens because the refractive index of air:
- remains constant with height
- decreases with height above the ground
- increases with height above the ground
- first increases then decreases with height
Answer: increases with height above the ground
In a desert, the ground heats the air above it, making the air near the surface hot and less dense (lower refractive index). Higher up, the air is cooler and denser (higher refractive index). Thus, refractive index increases with height. Light bends away from the normal as it goes from denser to rarer medium, and eventually TIR occurs.
7. A concave mirror (f = 20 cm) and a convex mirror (f = 30 cm) are placed coaxially 50 cm apart. An object is 30 cm left of the concave mirror. What is the nature of the final image?
- Virtual, 15 cm left of convex mirror
- Real, 15 cm left of convex mirror
- Real, 15 cm right of convex mirror
- Virtual, 15 cm right of convex mirror
Answer: Real, 15 cm left of convex mirror
For concave mirror: u1 = -30 cm, f1 = -20 cm → 1/v1 = 1/f1 - 1/u1 = -1/20 + 1/30 = -1/60 → v1 = -60 cm (real, left of concave). Image is 10 cm left of convex mirror, so u2 = +10 cm (virtual object). For convex mirror: f2 = +30 cm → 1/v2 = 1/30 - 1/10 = -1/15 → v2 = -15 cm. Negative v2 indicates real image, 15 cm left of convex mirror.
8. An object is placed 30 cm in front of a glass slab of thickness 10 cm and refractive index 1.5. Behind the slab is a spherical refracting surface of radius 20 cm separating glass (n = 1.5) from air (n = 1). What is the final image distance from the spherical surface? (Use paraxial approximation.)
- 60 cm
- 20 cm
- 40 cm
- 30 cm
Answer: 60 cm
The slab shifts the object by t(1 - 1/n) = 10(1 - 2/3) = 10/3 cm toward the surface. Effective object distance for the spherical surface = 30 - 10/3 = 80/3 cm. Using n2/v - n1/u = (n2 - n1)/R with n1 = 1.5, n2 = 1, u = -80/3 cm, R = -20 cm (convex toward object), we get 1/v - 1.5/(-80/3) = (1 - 1.5)/(-20). Solving gives v = 60 cm.
9. A ray of light passes through a parallel-sided glass slab. The emergent ray is parallel to the incident ray because:
- the slab faces are parallel and Snell's law gives i = e
- the angle of incidence equals the angle of emergence at both surfaces
- the refractive index of glass is constant
- the ray inside the slab is parallel to the incident ray
Answer: the slab faces are parallel and Snell's law gives i = e
For a parallel-sided slab, applying Snell's law at the first surface gives sin i = n sin r, and at the second surface n sin r = sin e. Since the faces are parallel, the angle of incidence at the second surface equals r. Combining gives sin i = sin e, so i = e. Hence the emergent ray is parallel to the incident ray.
10. A converging beam of light is incident on a concave mirror of focal length 20 cm. The beam would have converged to a point 30 cm behind the mirror if the mirror were absent. Where is the image formed after reflection?
- 12 cm behind the mirror
- 12 cm in front of the mirror
- 60 cm in front of the mirror
- 60 cm behind the mirror
Answer: 12 cm in front of the mirror
The converging beam would meet at a point 30 cm behind the mirror, so the object is virtual with u = +30 cm. For a concave mirror, f = -20 cm. Using mirror equation 1/v + 1/u = 1/f gives 1/v = 1/(-20) - 1/30 = -1/12, so v = -12 cm. Negative v indicates a real image 12 cm in front of the mirror.
11. Why does the sun appear flattened near the horizon at sunrise?
- Refraction lowers the upper edge more than the lower edge, stretching the vertical size
- Refraction raises the upper edge more than the lower edge, compressing the vertical size
- Refraction lowers the lower edge more than the upper edge, stretching the vertical size
- Refraction raises the lower edge more than the upper edge, compressing the vertical size
Answer: Refraction raises the lower edge more than the upper edge, compressing the vertical size
Atmospheric refraction bends light from the sun. Near the horizon, light from the lower edge travels through more atmosphere and bends more, raising its apparent position more than the upper edge. This differential refraction compresses the vertical size, making the sun appear flattened.
12. A prism of angle 60° and refractive index 1.5 is used. If the angle of incidence is 50°, what is the angle of deviation? (Given sin 50° = 0.766, sin 30.7° = 0.511, sin 29.3° = 0.489, sin 47.2° = 0.734)
- 30.0°
- 37.2°
- 40.0°
- 35.0°
Answer: 37.2°
Using Snell's law at first face: sin 50° = 1.5 sin r1 → r1 = arcsin(0.766/1.5) = arcsin(0.511) = 30.7°. Then r2 = A - r1 = 60° - 30.7° = 29.3°. At second face: 1.5 sin 29.3° = sin e → e = arcsin(1.5×0.489) = arcsin(0.734) = 47.2°. Finally δ = i + e - A = 50° + 47.2° - 60° = 37.2°.