In the Young's double slit experiment, the intensities at two points P₁ and P₂ on the screen are respectively I₁ and I₂. If P₁ is located at the centre of a bright fringe and P₂ is located at a distance equal to quarter of fringe width from P₁, then I₁ / I₂ is

$\mathrm{\Delta x}=\frac{\mathrm{dy}}{\mathrm{D}}=\frac{\mathrm{d}\left(\frac{\mathrm{\beta }}{4}\right)}{\mathrm{D}} \mathrm{and} \frac{\mathrm{\lambda }}{2\mathrm{\pi }}\mathrm{\Delta \varphi }=\frac{\mathrm{d}\left(\frac{\mathrm{\lambda D}}{\mathrm{d}}\right)}{4\mathrm{D}}$
$\mathrm{\Delta \varphi }=\mathrm{\pi }/2={\mathrm{\varphi }}_2 \mathrm{and} {\mathrm{\varphi }}_1=0$

Explanation:

Young's Double Slit Experiment: Intensity Ratio

Step 1: Understand the problem

• At P₁:
   

  P₁ is at the center of a bright fringe, so the phase difference φ₁ = 0.

  The intensity at P₁ is maximum:

  I₁ = I₀.

• At P₂:
   

  P₂ is located at a distance equal to a quarter of the fringe width from P₁.

  A quarter of the fringe width corresponds to a phase difference of φ₂ = π/2.

  The intensity at P₂ is:

  I₂ = I₀ cos²(φ₂/2).

Step 2: Calculate I₂

Substitute φ₂ = π/2:

I₂ = I₀ cos²(π/4).

The value of cos(π/4) is 1/√2:

I₂ = I₀ (1/√2)².

I₂ = I₀ × 1/2.

Step 3: Calculate I₁ / I₂

I₁ / I₂ = I₀ / (I₀ / 2).

I₁ / I₂ = 2.

Final Answer:

The ratio of intensities is:

I₁ / I₂ = 2.