Question

Diffraction pattern of single slit of width 0.5 cm
is formed by a lens of focal length 40 cm.
Calculate the distance between the first dark and
the next bright fringe from the axis. Wavelength
of light used is 4890 Ă.
(Ans: 1.956 x 10-2 mm)​

Answer 1

Answer:

Explanation:

=> It is given that,

width of single slit = 0.5 cm = 5 * 10^-3 m

focal length, f = 40 cm = 0.4 m

Wavelength of light, λ = 4890 Ă =4890 * 10^-10 m

=> For a minima:

a sinθ = nλ

but,  sinθ = x₁/f

Here, n = 1, we obtain x₁/f = λ/a

∴ x₁ =fλ/a

= 0.4*4890*10^-10 / 5*10^-3 m

= 3.912 * 10^-5 m.

=> For a maximam:

a sinθ = (2n+1)λ/2

but,  sinθ = x₂/f

Here, n = 1, so we obtain x₂/f = 3λ/2a

∴ x₂ =f3λ/2a

= 0.4*3*4890*10^-10 /2* 5*10^-3 m

= 5.868 * 10^-5 m.

=> Thus,distance between the first dark and  the next bright fringe:

x = x₂ - x₁

= 5.868*10^-5 - 3.912 * 10^-5 m

= 1.956 * 10^-5 m = 1.956 * 10^-2 mm

Thus, the distance between the first dark and  the next bright fringe from the axis is 1.956 * 10^-2 mm