MHT CET202527 Apr 2025Evening ShiftPhysicsWave OpticsActual
A ray of light is incident at polarising angle ' ' on air-glass interface, if ' _ a ' and ' _ g ' are the wavelengths of light in air and glass respectively, then
Options
- A_g= _a
- B_a= _g ^2
- C_g= _a ^2
- D_a= _g
Correct answer
A. _g= _a
Step-by-step solution
According to Brewster's law for light incident at the polarizing angle on an air-glass interface, = _g , where _g is the refractive index of glass relative to air. The refractive index may be expressed as _g = c / v_g , with c and v_g representing the speed of light in air and glass, respectively. Since frequency remains constant across media, c = f _a and v_g = f _g , leading to _g = _a / _g . Equating both expressions for _g gives = _a / _g , which rearranges to _g = _a . Final answer: A