Physics · Wave Optics · NEET
In Huygens theory, light travels as wavefronts made of secondary wavelets. When a wavefront goes from a denser medium (speed v1, small) to a rarer medium (speed v2, large), the part of the wavefront in the rarer medium moves faster and travels farther in the same time. This makes the refracted wavefront tilt away from the normal, so the refraction angle r is larger than the incidence angle i. As you keep increasing i, r increases faster. At one special angle (the critical angle ic), r becomes 90 degrees and the refracted wavefront lies flat on the surface. If i goes beyond ic, sin r would need to be greater than 1, which is impossible. So no refracted wavefront can be built in the rarer medium, and the entire wave is reflected back. That is total internal reflection.
The critical angle ic is the incidence angle at which the refracted wavefront becomes exactly parallel to the boundary surface, meaning the refraction angle r = 90 degrees. At this moment the refracted wave just grazes along the surface. From Snell's law written as n1 sin i = n2 sin r, put r = 90 so sin r = 1. This gives sin ic = n2 / n1 (going from denser medium n1 to rarer medium n2). NCERT writes this as sin ic = n2/n1. For any i larger than ic, the wavefront cannot enter the rarer medium at all.
Snell's law demands sin r = (n1/n2) sin i. Since n1 (denser) is greater than n2 (rarer), the factor n1/n2 is greater than 1. When i is bigger than ic, the right side becomes greater than 1. But sin r can never be more than 1 for a real angle. So there is no real refraction angle r, which means Huygens cannot construct a valid refracted wavefront in the rarer medium. With no wavefront to carry energy forward, all the energy stays in the denser medium as a reflected wave.
The light stays inside the same (denser) medium during TIR because it is reflected, not transmitted. So both the frequency and the wavelength stay exactly the same as before, and the speed is unchanged too. Frequency never changes at any boundary because it is fixed by the source. Since TIR keeps the wave in the original medium, nothing about the wave (speed, wavelength, frequency) changes; only its direction changes as it reflects.
No. TIR needs the light to go from a denser medium (higher refractive index, slower speed) to a rarer medium (lower index, faster speed). Only then does the refracted wavefront bend away from the normal and reach 90 degrees at a finite critical angle. When light goes from rarer to denser, the wavefront bends toward the normal, r is always smaller than i, and it can never reach 90 degrees. So a critical angle does not exist for rarer-to-denser travel, and TIR is impossible in that direction.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Light must travel from a denser to a rarer medium, and the angle of incidence must be greater than the critical angle. At the critical angle the refracted wavefront becomes parallel to the surface (r = 90 degrees); beyond it, no refracted wavefront can form, so the wave is fully reflected.
sin ic = n2 / n1, where n1 is the denser medium (light starts here) and n2 is the rarer medium. It comes from Snell's law n1 sin ic = n2 sin 90, since sin 90 = 1.
In the rarer medium the wave moves faster, so that part of the wavefront advances farther in the same time. This tilts the refracted wavefront. As incidence increases, the tilt grows until, at the critical angle, the refracted wavefront lies flat along the boundary, giving r = 90 degrees.
At exactly the critical angle the refracted wave just grazes along the surface at 90 degrees, carrying almost no energy into the rarer medium. Practically, essentially all the energy is reflected once the incidence angle reaches and exceeds ic.
No. In ideal total internal reflection, 100 percent of the light energy is reflected back into the denser medium. Because no wavefront enters the rarer medium, there is no transmitted (refracted) energy at all.
