AP Physics 2- 13.3 Refraction- Study Notes- New Syllabus
AP Physics 2- 13.3 Refraction – Study Notes
AP Physics 2- 13.3 Refraction – Study Notes – per latest Syllabus.
Key Concepts:
- Refraction of Light
- Snell’s Law
- Critical Angle (\(\theta_c\))
- Total Internal Reflection
Refraction of Light
Refraction is the bending of light as it passes from one medium to another due to a change in its speed.
The speed of light in a medium is less than in vacuum: \(\mathrm{v = \dfrac{c}{n}}\)
Refractive index (\(\mathrm{n}\)) of a medium is defined as:
\(\mathrm{n = \dfrac{c}{v}}\)
- \(\mathrm{c}\) = speed of light in vacuum (\(3 \times 10^8 \, m/s\))
- \(\mathrm{v}\) = speed of light in the medium
Snell’s Law:
Snell’s law describes the relationship between the angles of incidence and refraction when light passes between two media with refractive indices \(\mathrm{n_1}\) and \(\mathrm{n_2}\):
\(\mathrm{n_1 \sin \theta_1 = n_2 \sin \theta_2}\)
- \(\mathrm{\theta_1}\) = angle of incidence (measured from the normal)
- \(\mathrm{\theta_2}\) = angle of refraction (measured from the normal)
- \(\mathrm{n_1}\) = refractive index of the first medium
- \(\mathrm{n_2}\) = refractive index of the second medium
Bending of Light Ray:
- When light passes from a less dense medium (\(\mathrm{n_1 < n_2}\)) to a denser medium, it bends toward the normal.
- When light passes from a denser medium (\(\mathrm{n_1 > n_2}\)) to a less dense medium, it bends away from the normal.
- The amount of bending depends on the ratio of the refractive indices of the two media.
Applications of Refraction:
- Lenses (converging and diverging)
- Prisms and spectrometers
- Optical fibers (using total internal reflection)
- Apparent depth in water
Example:
A light ray passes from air (\(\mathrm{n_1 = 1}\)) into glass (\(\mathrm{n_2 = 1.5}\)) at an angle of incidence of 30°. Find the angle of refraction.
▶️ Answer/Explanation
Using Snell’s law: \(\mathrm{n_1 \sin \theta_1 = n_2 \sin \theta_2}\)
\(\mathrm{1 \cdot \sin 30^\circ = 1.5 \cdot \sin \theta_2}\)
\(\mathrm{\sin \theta_2 = \dfrac{0.5}{1.5} = 0.3333}\)
\(\mathrm{\theta_2 \approx 19.47^\circ}\)
Critical Angle (\(\theta_c\)):
The critical angle is the angle of incidence for which the angle of refraction is 90°:
\(\mathrm{\sin \theta_c = \dfrac{n_2}{n_1}}, \quad n_1 > n_2\)
- If \(\mathrm{\theta_i < \theta_c}\), refraction occurs.
- If \(\mathrm{\theta_i = \theta_c}\), the refracted ray grazes along the interface.
- If \(\mathrm{\theta_i > \theta_c}\), total internal reflection occurs.
Total Internal Reflection (TIR)
Total internal reflection occurs when a light ray traveling from a denser medium to a rarer medium hits the interface at an angle greater than the critical angle, causing the ray to be completely reflected back into the denser medium instead of refracting into the rarer medium.
Conditions for Total Internal Reflection:
- Light must travel from a denser medium (\(\mathrm{n_1}\)) to a rarer medium (\(\mathrm{n_2}\)) with \(\mathrm{n_1 > n_2}\).
- The angle of incidence (\(\mathrm{\theta_i}\)) must be greater than the critical angle (\(\mathrm{\theta_c}\)).
Properties of TIR:
- Occurs only at the interface from denser → rarer medium.
- The reflected ray obeys the law of reflection (angle of incidence = angle of reflection).
- There is no loss of light in an ideal case.
Applications of Total Internal Reflection:
- Optical fibers for telecommunications.
- Prisms in periscopes and binoculars (like Porro prisms).
- Endoscopes for medical imaging.
- Diamond sparkle is enhanced due to TIR inside the gem.
Example:
Light travels from water (\(\mathrm{n_1 = 1.33}\)) to air (\(\mathrm{n_2 = 1.0}\)). Find the critical angle.
▶️ Answer/Explanation
Using the critical angle formula:
\(\mathrm{\sin \theta_c = \dfrac{n_2}{n_1} = \dfrac{1.0}{1.33} \approx 0.7519}\)
\(\mathrm{\theta_c \approx 48.8^\circ}\)
Hence, any incidence angle greater than 48.8° will cause total internal reflection.