Refraction - AP Physics 2

A narrow light ray in air strikes a flat boundary with glass at an incident angle of $40^\circ$ measured from the normal. The glass has a higher refractive index than air ( $n_{glass} > n_{air}$ ). Ignoring reflection, the ray enters the glass at the point of incidence. Which way does the ray bend as it crosses into the glass?
Answer: It bends toward the normal because the speed decreases. This question tests understanding of refraction. When light travels from a medium with lower refractive index (air) to one with higher refractive index (glass), the light speed decreases because v = c/n, where n is the refractive index. At the boundary, the component of the wave's velocity parallel to the surface must remain continuous, while the perpendicular component changes due to the speed change. This causes the light ray to bend toward the normal (the line perpendicular to the surface) when entering a denser medium. Choice C incorrectly assumes that constant frequency prevents bending, but frequency conservation doesn't determine ray direction—the speed change does. When light slows down entering a denser medium, always remember it bends toward the normal.
A light ray travels in water and enters glass at an incident angle of $45^\circ$ to the normal. Glass has the higher refractive index ( $n_{glass} > n_{water}$ ). Which way does the ray bend as it enters the glass?
Answer: Toward the normal because the speed decreases. This question tests understanding of refraction. When light travels from water (lower refractive index) to glass (higher refractive index), the light speed decreases because v = c/n. At the boundary, maintaining the tangential velocity component while the normal component decreases causes the ray to bend toward the normal. The refracted ray in glass makes a smaller angle with the normal than the incident 45°. Choice D incorrectly claims no bending occurs because wavelength stays the same, but wavelength actually changes (λ = v/f) and bending is determined by the speed change, not wavelength. When light slows down entering a denser medium, it bends toward the normal.
A ray in air enters a liquid at $70^\circ$ to the normal. The liquid has the higher refractive index ( $n_{liquid} > n_{air}$ ). Compared to the incident ray, the refracted ray in the liquid is bent which way?
Answer: Toward the normal in the liquid This problem examines refraction. When light passes from air (lower refractive index) into a liquid (higher refractive index), the light wave slows down in the denser medium. This speed decrease causes the light ray to bend toward the normal at the boundary. Even with a large incident angle of 70°, the refracted ray bends toward the normal, resulting in a smaller angle with the normal in the liquid. Choice C incorrectly claims the ray bends away from the normal because speed decreases, showing confusion about how speed change relates to bending direction. Remember: when light enters a denser medium (higher n), it always slows down and bends toward the normal.
A monochromatic beam in air enters a slab of flint glass at $20^\circ$ to the normal. Flint glass has the higher refractive index ( $n_{flint} > n_{air}$ ). Which way does the ray bend upon entering the glass?
Answer: Toward the normal because the speed decreases. This question tests understanding of refraction. When light travels from air (lower refractive index) to flint glass (higher refractive index), the light speed decreases because v = c/n. At the boundary, maintaining the tangential velocity component while the normal component decreases causes the ray to bend toward the normal. The refracted ray in flint glass makes a smaller angle with the normal than the incident 20°. Choice C incorrectly claims frequency increases, but frequency is set by the source and remains constant across material boundaries—only speed and wavelength change. When light slows down entering a denser medium, it always bends toward the normal.
Light travels from glass into glycerin at an incident angle of $40^\circ$ to the normal. Glass has the higher refractive index ( $n_{\text{glass}} > n_{\text{glycerin}}$ ). The ray crosses the boundary into glycerin. Which way does the ray bend upon entering glycerin?
Answer: Away from the normal, because the speed increases in glycerin This question tests understanding of refraction. When light travels from glass (higher refractive index) to glycerin (lower refractive index), the light speed increases because v = c/n, where a lower n means higher speed. At the boundary, the wave component parallel to the interface maintains its speed, but the overall wave speed increase causes the ray to bend away from the normal. This behavior follows Snell's law: n₁sin(θ₁) = n₂sin(θ₂), where the smaller n₂ results in a larger refraction angle θ₂. Choice D incorrectly suggests frequency changes, but frequency remains constant across material boundaries—only speed and wavelength change with the medium. Remember: when light speeds up (entering a less dense medium), it bends away from the normal.
A ray of light passes from air into a clear plastic slab. The plastic has higher refractive index than air ( $n_{plastic} > n_{air}$ ). The incident angle is nonzero, and the ray enters the plastic. Which statement best describes the direction of the refracted ray?
Answer: It bends toward the normal because the speed decreases This question tests understanding of refraction. When light passes from air (lower refractive index) into plastic (higher refractive index), the wave speed decreases according to v = c/n. At the boundary, the portion of the wavefront entering the plastic first slows down while the rest continues at air speed, causing the entire wavefront to rotate toward the normal line. This bending toward the normal is characteristic of light entering a denser optical medium. Choice B incorrectly claims frequency increases, revealing the misconception that frequency changes during refraction—frequency depends solely on the source and remains constant. Remember: waves bend toward the normal when they slow down (entering higher n media).
A light ray in ethanol strikes an ethanol–glass interface at $25^\circ$ to the normal. Glass has the higher refractive index ( $n_{\text{glass}} > n_{\text{ethanol}}$ ). The ray enters the glass. Compared to the incident ray, how does the refracted ray bend?
Answer: Toward the normal, because the speed decreases in glass This question tests understanding of refraction. When light travels from ethanol (lower refractive index) to glass (higher refractive index), the light speed decreases since v = c/n and glass has a larger n. At the boundary, the wave's parallel component maintains continuity, but the overall speed decrease causes the ray to bend toward the normal. This follows Snell's law: n₁sin(θ₁) = n₂sin(θ₂), where the larger n₂ (glass) requires a smaller angle θ₂ from the normal. Choice D incorrectly claims frequency changes, but frequency is an intrinsic property of the light source and remains constant during refraction. The key principle is: when light slows down (entering a denser medium), it always bends toward the normal.
A laser beam travels in glass and reaches a flat glass–air boundary at $20^\circ$ to the normal. Glass has a higher refractive index than air. Neglect reflection. Compared to the incident ray, the refracted ray in air bends which way?
Answer: Away from the normal, because the speed increases in air This question tests understanding of refraction. When light travels from a medium with higher refractive index (glass) to one with lower refractive index (air), the light speed increases. As the wave fronts cross the boundary, those entering air first travel faster than those still in glass, causing the ray to bend away from the normal. Choice D incorrectly claims wavelength stays constant—wavelength actually increases in air while frequency remains constant, revealing the misconception that wavelength is conserved across boundaries. Remember: when light enters a less dense medium (lower n), it bends away from the normal.
A light ray travels from glass into water, striking the boundary at $30^\circ$ to the normal. Glass has the higher refractive index ( $n_{glass} > n_{water}$ ), and the ray transmits into the water. Compared to the incident angle, the refracted angle in water is ____.
Answer: larger because the speed increases This question tests understanding of refraction. When light travels from glass (higher refractive index) to water (lower refractive index), the light speed increases since v = c/n. At the boundary, the parallel velocity component stays constant while the perpendicular component increases, causing the ray to bend away from the normal. Therefore, the refracted angle in water is larger than the incident angle of 30°. Choice B incorrectly states frequency decreases, but frequency remains constant across boundaries—it's determined by the source and doesn't change with the medium. When light speeds up entering a less dense medium, remember it bends away from the normal.
A ray travels in water and refracts into air at a smooth, flat boundary. Water has the higher refractive index ( $n_{water} > n_{air}$ ). If the ray does refract into air, which way is it bent relative to the normal?
Answer: Away from the normal because the speed increases This question tests understanding of refraction. When light travels from water (higher refractive index) to air (lower refractive index), the wave speed increases according to v = c/n, since n_air < n_water. As the wavefront crosses the boundary, the part entering air first speeds up while the remainder stays at water speed, causing the wavefront to rotate away from the normal. This bending away from the normal always occurs when light enters a less dense optical medium. Choice D incorrectly claims frequency increases, showing the misconception that frequency varies with medium—frequency depends only on the source and never changes during refraction. Remember: when light speeds up (entering lower n media), it bends away from the normal.
A narrow light ray in air strikes a flat boundary with water at an incident angle of $35^\circ$ to the normal. Water has a higher refractive index than air. Ignore reflection and assume the boundary is smooth. Which way does the ray bend as it enters the water?
Answer: Toward the normal, because the speed decreases in water This question tests understanding of refraction. When light travels from a medium with lower refractive index (air) to one with higher refractive index (water), the light speed decreases. At the boundary, the wave fronts on the water side travel slower than those still in air, causing the ray to bend toward the normal—like a car wheel hitting sand at an angle. The frequency of light never changes when crossing boundaries, making choice C incorrect based on the misconception that frequency determines bending direction. Remember: when light enters a denser medium (higher n), it bends toward the normal.
A light ray in acrylic strikes a boundary with air at an incident angle of $15^\circ$ measured from the normal. Acrylic has the higher refractive index ( $n_{acrylic} > n_{air}$ ), and the ray refracts into the air. Which way does the transmitted ray bend at the boundary?
Answer: Away from the normal because the speed increases. This question tests understanding of refraction. When light travels from acrylic (higher refractive index) to air (lower refractive index), the light speed increases because v = c/n. At the boundary, maintaining continuity of the tangential velocity component while the normal component increases causes the ray to bend away from the normal. The transmitted ray makes a larger angle with the normal than the incident 15°. Choice C incorrectly suggests frequency increases, but frequency is invariant across material boundaries—only speed and wavelength change with the medium. Remember: when light enters a less dense medium and speeds up, it bends away from the normal.
A ray of light travels in air and enters water at an incident angle of $50^\circ$ to the normal. Water has the higher refractive index ( $n_{\text{water}} > n_{\text{air}}$ ). Neglecting reflection, which way does the transmitted ray bend in the water?
Answer: Toward the normal, because the speed decreases in water This question tests understanding of refraction. When light travels from air (lower refractive index) to water (higher refractive index), the light speed decreases because v = c/n, where water's higher n means lower speed. At the air-water boundary, the wave component parallel to the interface must remain continuous, but the overall speed reduction causes the ray to bend toward the normal. This behavior is described by Snell's law: n₁sin(θ₁) = n₂sin(θ₂), where water's larger n₂ requires a smaller refraction angle θ₂. Choice C incorrectly suggests frequency increases, but frequency is determined by the source and never changes during refraction. The fundamental rule is: waves bend toward the normal when they slow down (entering a denser medium).
A light ray travels from air into crown glass at $10^\circ$ to the normal. Crown glass has the higher refractive index ( $n_{\text{glass}} > n_{\text{air}}$ ). Which statement correctly describes the ray's bending at the boundary?
Answer: Toward the normal, because the speed decreases in glass This question tests understanding of refraction. When light travels from air (lower refractive index) to crown glass (higher refractive index), the light speed decreases because v = c/n, where glass's higher n results in lower speed. At the air-glass boundary, the wave's parallel component maintains continuity, but the overall speed reduction causes the ray to bend toward the normal. This is described by Snell's law: n₁sin(θ₁) = n₂sin(θ₂), where glass's larger n₂ requires a smaller refraction angle θ₂. Choice C incorrectly suggests frequency increases, but frequency is an invariant property during refraction—only wavelength and speed change with the medium. Remember: waves bend toward the normal when they slow down (entering a denser medium).
A light ray in diamond reaches a flat diamond–water boundary at $10^\circ$ to the normal. Diamond has a higher refractive index than water. Neglect reflection. Compared to the incident ray, the refracted ray in water bends which way?
Answer: Away from the normal, because the speed increases in water This question tests understanding of refraction. When light travels from diamond (higher refractive index) to water (lower refractive index), the light speed increases. As the wave fronts enter water, they speed up relative to those still in diamond, causing the ray to bend away from the normal. Choice C incorrectly attributes the bending to frequency decrease, revealing the misconception that frequency changes at boundaries—frequency actually remains constant while wavelength increases. Remember: when light exits into a less dense medium, it speeds up and bends away from the normal.
A light ray in air enters a transparent plastic slab at $55^\circ$ to the normal. The plastic has the higher refractive index ( $n_{plastic} > n_{air}$ ). Which statement best describes the ray's direction after refraction?
Answer: It bends toward the normal in the plastic This problem examines refraction. When light passes from air (lower refractive index) into plastic (higher refractive index), the light wave slows down in the denser medium. This decrease in speed causes the light ray to bend toward the normal at the boundary. The refracted ray in plastic makes a smaller angle with the normal than the 55° incident angle. Choice C incorrectly claims the speed increases in plastic, demonstrating the common misconception that confuses the relationship between refractive index and wave speed. To solve refraction problems: identify which medium has higher n, then apply the rule that light bends toward the normal when entering a denser medium.
A laser beam travels in water and reaches a water–air boundary at $20^\circ$ to the normal. Water has the higher refractive index ( $n_{\text{water}} > n_{\text{air}}$ ). The beam refracts into air. Compared to the incident ray, which way does the refracted ray bend?
Answer: Away from the normal, because the speed increases in air This question tests understanding of refraction. When light travels from a medium with higher refractive index (water) to one with lower refractive index (air), the light speed increases because v = c/n. As the wave crosses the boundary, the parallel component of velocity must remain continuous, but since the overall speed increases in air, the wave must bend away from the normal to accommodate this speed change. This follows from Snell's law: n₁sin(θ₁) = n₂sin(θ₂), where the smaller n₂ (air) requires a larger θ₂ (angle from normal). Choice C incorrectly claims frequency changes, but frequency is determined by the source and remains constant during refraction. The transferable principle is: when light enters a medium with lower refractive index (faster speed), it always bends away from the normal.
A laser beam travels in water and strikes a boundary with air at $25^\circ$ to the normal. Water has the higher refractive index ( $n_{water} > n_{air}$ ). Which way does the refracted ray bend as it enters the air?
Answer: Away from the normal in the air This question involves refraction. When light travels from a medium with higher refractive index (water) to one with lower refractive index (air), the light wave speeds up as it enters the less dense medium. This speed increase at the boundary causes the light ray to bend away from the normal line. The refracted ray makes a larger angle with the normal than the incident ray because the wavefront accelerates as it crosses into air. Choice D incorrectly suggests frequency decreases, showing the misconception that frequency changes during refraction when actually frequency remains constant. The key principle: when light enters a less dense medium (lower n), it speeds up and bends away from the normal.
A beam travels in air and enters water at $35^\circ$ to the normal. Water has the higher refractive index ( $n_{water} > n_{air}$ ). Which option correctly describes the refraction direction at the boundary?
Answer: It bends toward the normal in water This problem addresses refraction. When light travels from air (lower refractive index) to water (higher refractive index), the light wave slows down as it enters the denser medium. This speed reduction at the boundary causes the light ray to bend toward the normal. The refracted ray in water makes a smaller angle with the normal than the 35° incident angle. Choice C incorrectly states the ray bends away from the normal because speed decreases, revealing confusion about the relationship between speed change and bending direction. To predict refraction: when light enters a denser medium (higher n), it slows down and bends toward the normal.
A ray in air strikes a boundary with ethanol at $35^\circ$ to the normal and refracts into the ethanol. Ethanol has the higher refractive index ( $n_{ethanol} > n_{air}$ ). Which statement correctly describes the ray's path in ethanol?
Answer: It bends toward the normal because the speed decreases. This question tests understanding of refraction. When light passes from air (lower refractive index) into ethanol (higher refractive index), the light speed decreases according to v = c/n. At the boundary, the tangential velocity component remains continuous while the normal component decreases, causing the ray to bend toward the normal. The refracted ray in ethanol makes a smaller angle with the normal than the incident 35°. Choice D incorrectly suggests constant frequency prevents bending, but frequency conservation doesn't determine ray direction—the speed change at the boundary causes the bending. When light slows down entering a denser medium, always remember it bends toward the normal.

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