Question

A beam of light consisting of two wavelengths 500 nm and 400 nm is used to obtain interference fringes in youngs double slit experiment. The distance between the slits is 0.3 mm and the distance between the slits and the screen is 1.5 m. Compute the least distance of the point from the central maximum, where the bright fringes due to both the wavelengths coincide

Answer 1

Answer:

Explanation:

Given A beam of light consisting of two wavelengths 500 nm and 400 nm is used to obtain interference fringes in youngs double slit experiment. The distance between the slits is 0.3 mm and the distance between the slits and the screen is 1.5 m. Compute the least distance of the point from the central maximum, where the bright fringes due to both the wavelengths coincide

Let the relative velocity of the particle with respect to hemisphere be Vr and v be the linear velocity. By conservation of linear momentum we get

5mv = m(Vr cosθ – v)

5mv = mVr cosθ – mv

5mv + mv = mVrcosθ

6mv = mVrcosθ

6v = Vr cosθ

Vr = 6v / cosθ

 ω = Vr / R = 6v / R cosθ

Given D = 1.5 m, d= 0.3 mm = 0.3 x 10^-3 m, λ1 = 500 x 10^-9 m, λ2 = 400 x 10^-9 m

Y = n1 λ1D/d = n2 λ2D/d

  n1 λ1 = n2 λ2

500 x 10^-9 n1 = 400 x 10^-9 n2

5 n1 = 4n2

Since n are integers we get n1 = 4 and n2 = 5

Now Y = n1 λ1D /d

   Y = 4 x 500 x 10^-9 x 1.5 / 0.3 x 10^-3

Y = 4 x 500 x 10^-9 x 5 x 10^3

Y = 2000 x 5 x 10^-6

Y = 10^-2  

Y = 0.01 m

Y =1 cm