ai. Why are parabolic mirrors suitable for use in the headlamps of vehicles?
ii. Draw a ray diagram to illustrate the answer in 10(a)(i).
bi. State two applications of echoes.
ii. An observer standing at a point, P, on the same horizontal ground as the foot, H, of a tower, shouts, and 1.20 s later, he hears the echo. He then moved to another point, Q, 40 m from P, and shouted again but the echo was heard after 1.45 s. Calculate the:
I. distance between P and H;
II. speed of sound in air.
ci. Define the term absolute refractive index of a medium.
ii. A piece of coin falls accidentally into a tank containing two immiscible liquids A and B as illustrated in Fig. 10.0 above.
Calculate the displacement of the coin when viewed vertically from above.
[refractive index of A = 1.3, refractive index of B = 1.4]
Explanation
ai. Parabolic mirrors, due to their ability to efficiently collect and focus incoming parallel light rays, are suitable for use in vehicle headlamps, ensuring a concentrated and directed beam of light for improved visibility and reduced glare.
ii. 
bi. – Sonar Systems: Sonar (Sound Navigation and Ranging) systems use echoes to determine the depth and location of objects underwater. A sound signal is emitted, and the time it takes for the signal to bounce off an object and return as an echo is used to calculate the distance to the object. Sonar is commonly used in marine navigation, fishing, and underwater mapping.
- Medical Imaging: In medical ultrasound imaging, echoes are used to create images of the interior of the body. High-frequency sound waves are directed into the body, and the echoes that bounce back from different tissues are used to generate detailed images of organs, blood vessels, and developing fetuses during pregnancy.
- Architectural Design: Architects and engineers use echoes to evaluate the acoustics of buildings and design spaces with optimal sound qualities. The controlled use of echoes can enhance the sound in concert halls, auditoriums, and theaters, providing a more enjoyable listening experience for audiences.
- Geological Surveys: Seismic surveys use echoes generated by controlled explosions or mechanical sources to map subsurface geological features. By analyzing the time it takes for seismic waves to reflect off different rock layers, geologists can infer the composition and structure of the Earth's subsurface.
- Surveying and Distance Measurement: Echoes are used in surveying and distance measurement equipment like total stations and laser rangefinders. A laser or light signal is emitted, and the time it takes for the light to bounce off a target and return is used to measure distances accurately.
- Echo Sounding in Navigation: In addition to sonar, echo sounding is used in navigation to determine water depth, especially in shallow or coastal areas. A sound signal is sent to the seabed, and the time it takes for the echo to return helps sailors and navigators avoid underwater hazards.
- Musical Effects: Musicians and audio engineers often use echoes and reverb effects in music production. These effects add depth and dimension to music by simulating the reflection of sound in different environments. Echoes and reverb are commonly used in recording studios, live concerts, and digital audio effects.
ii. V = \(\frac{2d}{t}\)
Case1: V = \(\frac{2d}{1.2}\)
Case2: V = \(\frac{2(40 + d)}{1.45}\)
Equating case 1 and 2
\(\frac{2d}{1.2}\) = \(\frac{2(40 + d)}{1.45}\)
2.9d = 96 + 2.4d
2.9d - 2.4d = 96
0.5d = 96
d = \(\frac{96}{0.5}\)
= 192m
∴ Distance between P and H = 192m
put d = 192m into case(1)
V = \(\frac{2d}{1.2}\)
V = \(\frac{2\times192}{1.2}\) = 320m/s
ci. The absolute refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in the medium. It is a measure of how much a medium slows down light.
ii. Displacement = real depth - \(\frac{real depth}{refractive index }\) ( n = \(\frac{real depth}{apparentdepth }\))
= real depth( 1 - \(\frac{1}{n}\))
For liquid A, displacement,d = 8 (1 - \(\frac{1}{1.3}\)) = 9.23cm
For liquid B, displacement,d = 40( 1 - \(\frac{1}{1.4}\)) = 2.29cm
∴ Total displacement = 9.23 + 2.29 = 11.52cm
∴ the apparent position of the coin is 11.52cm from the bottom.