Question

3. A thin layer of turpentine (n-1.472) is floating on water (-1.333). Light of wavelength 589 nm initially traveling in air is incident on the turpentine at an angle of 24.8° measured with respect to the air turpentine interface. b. c. After the light passes through the turpentine, what is its refraction angle in the water? At what angle should the light be incident on the turpentine-water interface so that all the light totally internally reflects in the turpentine and does not travel into the water?

Answer 1

6529 249 laip Using Shelt law Sinc h = 1, haz 1.472 i = 65.2° yo sing - (6 sino = hi sini = I sin 1652) 1.472 sino = 0.6162 8= sint(0.6167) (8= 38.08° Answer Similarly CAt surface 2) 2 h32= hz - sine Here i= 0 = 38.080 - Sin ginf=h2 sino И - 1.472 sin(38.089) 1.333 0.68108 Tz = sin(0.68108) - 42.939 (8= 42.930) Incident angle at the fate of interface is should be greader than critical angle ie= sin(n ) = sen-1 (1333) = 64.90° So i > 64.9°/ for total internal reflection, ②