Question

Explain Young’s double-slit experimental setup and obtain the equation for path difference.

Answer 1

Young’s double-slit experiment

1. Wavefronts from S₁ and S₂ spread out and overlapping takes place to the right side of the double slit.
2. When a screen is placed at a distance of about 1 meter from the slits, alternately bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.
   Equation for path difference:
3. Let d be the distance between the double slits S₁ and S₂ which act as coherent sources of wavelength λ.
4. A screen is placed parallel to the double-slit at a distance D from it. The mid-point of S₁ and S₂ is C and the mid-point of the screen O is equidistant from S₁ and S₂. P is any point at a distance y from O.
5. The waves from S₁ and S₂ meet at P either in-phase or out-of-phase depending upon the path difference between the two waves.
   
   Young’s double-slit experimental setup

   δ = S₂P – S₁P
   δ = S₂P – MP = S₂M

6. The angular position of the point P from C is θ. ∠ OCP = θ.

7. From the geometry, the angles ∠ OCP and ∠ S₂S₁M are equal.
   ∠ OCP = ∠ S₂S₁M = θ.
   In right-angle triangle ∆ S₁S₂M, the path difference,
   S₂M = d sin θ
   δ = d sin θ
   If the angle θ is small, sin θ »tan θ θ
   From the right angle triangle ∆ OCP,
   tan θ = `"y"/"D"`
   The path difference, δ = `"dy"/"d"`