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I) The "sheet of water" seen on a hot day is an optical illusion called a mirage. It is caused by the refraction and total internal reflection of light. On a hot day, the air near the ground i

Physics

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Answer

b(ii)

127.5 m\text{127.5 m}

Step-by-step solution

3 steps

Part a(ii)

The cause of the mirage is the phenomenon of refraction and total internal reflection. Applications of these phenomena include

i) The "sheet of water" seen on a hot day is an optical illusion called a mirage. It is caused by the refraction and total internal reflection of light. On a hot day, the air near the ground is much hotter and less dense than the air higher up. Light rays from the sky (or distant objects) travel from the cooler, denser air to the hotter, rarer air near the ground. As light passes through layers of air with different densities, it bends. When the angle of incidence of light from the sky on the boundary between air layers exceeds the critical angle, total internal reflection occurs, causing the light to bend upwards into the observer's eyes. This makes the sky appear to be on the ground, resembling a pool of water.

  • Optical fibres: Used in telecommunications and medical endoscopes to transmit light signals over long distances with minimal loss, relying on total internal reflection.
  • Lenses: Used in spectacles, cameras, telescopes, and microscopes to focus or diverge light through refraction.
  • Prisms: Used in binoculars and periscopes to change the direction of light through total internal reflection, and in spectroscopes to disperse light through refraction.

Part b(ii)

127.5 m\text{127.5 m}

To estimate the distance between the buildings, we use the formula

i) The sound heard from the locked building is an echo. An echo occurs when sound waves produced by a source travel through a medium, strike a distant surface (like another building), and are reflected back to the listener. The reflected sound is perceived as a distinct repetition of the original sound if the time delay between the original sound and the reflected sound is sufficient.

Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}

The time given (0.75 s) is the time taken for the sound to travel from the locked building to the reflecting building and then back to the observer. Therefore, the total distance travelled by the sound is twice the distance between the buildings.

We assume the speed of sound in air is approximately 340 m/s340 \text{ m/s}.

  1. Identify the given values.

    Time, t=0.75 st = 0.75 \text{ s}
    Speed of sound, v=340 m/sv = 340 \text{ m/s}

  2. Calculate the total distance travelled by the sound.

    Total distance =v×t= v \times t
    Total distance =340 m/s×0.75 s= 340 \text{ m/s} \times 0.75 \text{ s}
    Total distance =255 m= 255 \text{ m}

  3. Calculate the distance between the buildings.

    Since the sound travels to the building and back, the distance between the buildings is half of the total distance travelled.
    Distance between buildings, d=Total distance2d = \frac{\text{Total distance}}{2}
    d=255 m2d = \frac{255 \text{ m}}{2}
    d=127.5 md = 127.5 \text{ m}

    The estimated distance between the buildings is 127.5 m\boxed{\text{127.5 m}}.

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