CHRISTIAAN HUYGENS (1629 – 1695)
Chapter 2: Chapter 10 · PHYSICS PART-2 · EN medium
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Christiaan Huygens ( – ) Dutch physicist, astronomer, mathematician and the founder of the wave theory of light. His book, Treatise on light, makes fascinating reading even today. He brilliantly explained the double refraction shown by the mineral calcite in this work in addition to reflection and refraction. He was the first to analyse circular and simple harmonic motion and designed and built improved clocks and telescopes. He discovered the true geometry of Saturn’s rings. Thus we obtain sin sin i v r v ( . ) From the above equation, we get the important result that if r < i (i.e.
📖 NCERT Class 12 Physics Part 2 · Page 51
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Christiaan Huygens ( – ) Dutch physicist, astronomer, mathematician and the founder of the wave theory of light. His book, Treatise on light, makes fascinating reading even today. He brilliantly explained the double refraction shown by the mineral calcite in this work in addition to reflection and refraction. He was the first to analyse circular and simple harmonic motion and designed and built improved clocks and telescopes.
He discovered the true geometry of Saturn’s rings. Thus we obtain sin sin i v r v ( . ) From the above equation, we get the important result that if r < i (i.e., if the ray bends toward the normal), the speed of the light wave in the second medium (v2) will be less then the speed of the light wave in the first medium (v1). This prediction is opposite to the prediction from the corpuscular model of light and as later experiments showed, the prediction of the wave theory is correct.
Now, if c represents the speed of light in vacuum, then, c v ( . ) and n2 = c v ( . ) are known as the refractive indices of medium and medium , respectively. In terms of the refractive indices, Eq.
( . ) can be written as n1 sin i = n2 sin r ( . ) This is the Snell’s law of refraction. Further, if # and # denote the wavelengths of light in medium and medium , respectively and if the distance BC is equal to # then the distance AE will be equal to # (because if the crest from B has reached C in time *, then the crest from A should have also reached E in time * ); thus,
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