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Explain the activity to understand the blue color of the sky and the reddish appearance of the sunrise and sunset.

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Answer
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Hint: The main thing you have to possess in order to answer this question is the clarity in the concepts of reason behind the blue color of the sky and its reddish appearance at sunrise and sunset. Thus you could design some activity that would clearly explain how these colors appear in the sky.

Complete answer:
In the question we are asked about the blue color of the sky and also the reddish appearance of the sunrise and sunset. As a physicist one should be aware of the fact that everything has a scientific explanation. We are asked to find the activity to understand the given two things.
One could place a light source S like an electric bulb at a converging lens’s focus so as to get a parallel beam of light. Now on passing this light through a transparent glass beaker of say, 500mL containing about 400mL of water the light can be allowed to pass through a circular hole on a cardboard C and thus obtain a sharp image of the hole on screen MN using another lens (convex). You could then place a beaker B with 200g sodium thiosulphate solution. Then, you could add 1 to 2mL concentrated sulphuric acid dropwise.
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OBSERVATION:
Microscopic sulfur particles would now precipitate and thus you will be able to observe the blue light on the sides as the result of scattering of short wavelengths by the minute particles. Now the color observed on the screen would change from orange red color to crimson red color.

Note: Though we have given the activity to show we understand the phenomenon given in the question, we haven’t really explained the reason behind it which would be scattering of light. The small particles present in the atmosphere scatter sunlight and blue light with shorter wavelength scatters the most and hence the blue color of the sky during daytime. Now during the sunset and sunrise the sunlight has to travel greater distance to reach our eyes and by the time it reaches only the red light with greater wavelength remains.