InterviewSolution
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InterviewSolution
| 1. |
ProveBrewster's Law. |
Answer» Solution :As an unpolarised light enters the denser medium, a part of light at the point of incidence gets reflected and the rest refracted in the dense medium. It is noticed that the refracted beam of light is either partially polarised or COMPLETELY UNPOLARIZED. However, the reflected beam gets completely polarised in the plane at right ANGLES to the plane ofincidence at a particular angle of incidence called polarising angles. It is also noticed that the angle between the completely polarised reflected beam and the refracted beam are at right angles to each other. Applying Snell's law of reflection, we write absolute refractive index, (rarer medium is taken as air) `mu=(sin i_B)/(sin r)""...(1)` where `i_B-` polarising angle or Brewster's angle. Also, `BOC=90^@` i.e., `NON'=180^@=NOB+BOC+N'OC` i.e., `180^@=i_B+90^@+r` `therefore i_B+r=90^@""....(2)` Angle of refraction `r=90^@-i_B""...(3)` USING (3) in (1) we get `mu=(sin i_B)/(sin (90^@-i_B))` `therefore mu=(sin i_B)/(cos i_B)` or `tan i_B=mu""...(4)` Expression (4) is known as the Brewster's law. The tangent of polarising angle for air-medium interface is equal to the absoluate refractive index of the medium for other than air-medium interface, `tan i_B=mu_2=(mu_2)/(mu_1)` where `mu_2=(c )/(v_2)` and `mu_1=(c )/(v_1) and mu_2 gt mu_1, v_1 and v_2` represent speed of light in the two OPTICAL media (1) and (2). Note: David Brewster, in the year 1811, discovered that at the polarising angle of incidence, the reflected light is completely plane polarised. |
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