Question

a) Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a maximum only for 487.9 nm and 683.0 nm in the visible spectrum. What is the minimum thickness of the film (n=1.58)?

t = _________ nm

b) A planoconvex lucite lens 3.6 cm in diameter is placed on a flat piece of glass as in the figure 34-18 in the textbook. When 600 −nm light is incident normally, 47 bright rings are observed, the last one right at the edge.

What is the radius of curvature of the lens surface?

R = ___________ m

What is the focal length of the lens?

f = ____________ m

Answer 1

wavelength λ1 =487.9 nm and λ2 =683 nm

[(m₁+1/2)/(m₂+1/2)] =(683 nm)/(487.9 nm)

      [(m₁+1/2)/(m₂+1/2)] = 1.4

      [(m₁+1/2)/(m₂+1/2)] = 7/5

     [(2m₁+1)/(2m₂+1)] = 7/5 ......... (4)

from eq (4),

   (2m₁+1) = 7   and (2m₂+1) = 5

therefore, m₁ = 3 and m₂ =2

substitute the m₁ value in eq (2), we get

              2t(1.58) =(3+1/2)(487.9 nm)   

            thickness t = 540.39 nm = 540.39*10^-9 m

B)

  r = √[(m+0.5)λ.R]

Here, r = radius of the lens = 3.6 cm/2 = 1.8 cm

          m =47 ; λ = 600 nm;

Radius of curvature of the lens,

           R = r²/[(m+0.5)λ] = (1.80*10^-2)²/[47.5*600*10^-9]

R = 11.36 m

Focal length,

          1/f =(n-1)/R ; n for lucyte is 1.495

            f = R/(n-1) = 11.36/0.495

f = 22.92 m