Question

A piece of blue glass heated to a high temperature and a piece of red glass at room temperature are taken inside a dimly lit room, then:
A. The blue piece will look blue and red will look as usual
B. Red looks brighter red and blue looks ordinary blue
C. Blue shines like brighter red compared to the red piece
D. Both the pieces will look equally red

Answer 1

Stefan-Boltzmann law describes the relation between heat emitted and temperature. Use Stefan-Boltzmann law statement to answer how the heated blue and red glasses appear after taken inside a dimly lit room.

Complete step by step answer: We are given that a piece of blue glass heated to a high temperature and a piece of red glass at room temperature.
We have to find how these glass pieces appear after taking them inside a dimly lit room.
Stefan-Boltzmann law states that the radiant heat power emitted by a body is directly proportional to the fourth power of temperature of the body.
Here the blue glass piece is heated at higher temperature compared to the red one.
So the blue glass absorbs all the radiation of higher wavelength except blue. So when it is taken in a dark room, it emits the radiation it absorbed.
Therefore, it appears brighter red than the red glass piece.
Hence, the correct option is Option C, Blue shines like brighter red compared to the red piece.

Note: Red reflects red, green reflects green, blue reflects blue and so on. When objects of some wavelength are heated then those objects absorb radiation of higher wavelengths and when some wavelength falls upon the same wavelength, then the light reflects. Here red wavelength is higher than blue wavelength. So blue absorbs red and appears redder. We perceive the color of any object because that object reflects the light of its color only. Blue becomes and appears red after heating. So be careful with the emission and absorption.