A green laser source (590 nm) of 1 W power is contained within a water bath (ref

Question

A green laser source (590 nm) of 1 W power is contained within a water bath (refractive index n=1.3). The top surface of the bath is open to air, whereas the bottom surface is bounded by a thick layer of glass (refractive index n=1.65), as shown in the diagram below. The source is polarized such that it is TM polarized with respect to the interfaces.

a)At what angle will the light from the source (S) be completely reflected from the water-air interface?

b)At what angle 1 will there be a light component measured at point 2, but none measured at point 1?

c)At this incident angle, how much light power will be measured at point 2?

Explanation / Answer

A) condition for total internal reflection is

theta1 = sin^(-1)(n1/n2) = sin^(-1)(1/1.3) = 50.3 deg

B) angle of incidence for water-glass interface

theta_1 = sin^(-1)(1.3/1.65) = 52 deg

C) Light power doesnot changes ,so the answer is 1 W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.