Question
Question: A polished silvery surface is a: A. Good absorber good reflector of heat B. Good absorber bad ra...
A polished silvery surface is a:
A. Good absorber good reflector of heat
B. Good absorber bad radiator of heat
C. Poor absorber and good reflector of heat
D. Poor reflector and good radiator of heat
Solution
In this question we have been asked about the nature of a polished silver surface. We have been given statements about the properties such as reflection, absorption and radiation. We have been asked to select the correct statement regarding the polished silver surface. Therefore, we shall briefly discuss the properties mentioned in the options. We shall later determine the nature of the surface.
Step by step solution:
We know that a polished silvery surface is a good reflector of light. The mirror is an excellent example of polished silver surface. We know that light rays from mirrors are perfectly reflected provided that the mirror is highly polished and smooth.
We know that reflection is the property of a material because of which it throws back the waves after they hit the surface. Similarly, absorption is the property by which the material absorbs all the incident rays on it. The property of radiation is the exact opposite of absorption, in that the body radiates energy or can be said to be the source of the waves.
Now, we know that a polished mirror is a good reflector. We also know that the black bodies are the best radiator and absorber. A mirror however is said to be the white body. The mirror does not very well absorb the heat waves since it reflects them. The mirror can not generate heat waves therefore the mirror is not a good radiator as well. Therefore, a mirror is a good reflector and a poor absorber as well as a poor radiator.
Therefore, the correct answer is option C.
Note:
When the light rays bounce off an object or a surface the phenomenon is said to be reflection. If the mirror is highly polished and smooth the angle of reflection will be equal to the angle of incidence meaning that the angle at which the light or het waves will bounce back will be equal to the angle at which the waves were incident on the surface.