In an arrangement of double slit experiment the slits are separated by 0.25cm.An interference pattern is formed on the screen 120cm away from the slits which are illuminated by a coherent light of wavelength of 600nm .The distance from the central maximum for which the intensity on the screen is 75.0% of the maximum is

To find the distance from the central maximum where the intensity is 75.0% of the maximum in a double slit experiment, we can use the principles of interference and the formula for intensity in terms of path difference. Let's break this down step by step.

Understanding the Setup

In the double slit experiment, coherent light passes through two closely spaced slits, creating an interference pattern on a screen. The key parameters in your scenario are:

• Separation between the slits (d): 0.25 cm = 0.0025 m
• Distance from the slits to the screen (L): 120 cm = 1.2 m
• Wavelength of the light (λ): 600 nm = 600 x 10^-9 m

Intensity Formula in Interference

The intensity of light at a point on the screen in a double slit experiment can be expressed as:

I = I_max * cos²(Δφ/2)

Where:

• I_max is the maximum intensity.
• Δφ is the phase difference between the two waves arriving at that point.

Calculating the Path Difference

The path difference (Δ) between the two waves reaching a point on the screen can be calculated using:

Δ = d * sin(θ)

For small angles, sin(θ) can be approximated as:

sin(θ) ≈ tan(θ) = y / L

Where y is the distance from the central maximum to the point of interest on the screen. Thus, we can rewrite the path difference as:

Δ = d * (y / L)

Finding the Phase Difference

The phase difference (Δφ) is related to the path difference by:

Δφ = (2π/λ) * Δ

Substituting the expression for Δ, we get:

Δφ = (2π/λ) * (d * (y / L))

Setting Up the Intensity Condition

We want to find the distance y where the intensity is 75% of the maximum intensity:

I = 0.75 * I_max

Using the intensity formula, we set:

0.75 * I_max = I_max * cos²(Δφ/2)

This simplifies to:

cos²(Δφ/2) = 0.75

Taking the square root gives:

cos(Δφ/2) = √0.75 = 0.866

Finding the Angle

Now, we can find Δφ/2:

Δφ/2 = cos^-1(0.866)

This gives us:

Δφ/2 ≈ 30° (since cos(30°) = √3/2 ≈ 0.866)

Thus, we have:

Δφ = 60° = (2π/3) radians

Relating Back to Path Difference

Now we can relate this back to the path difference:

Δφ = (2π/λ) * (d * (y / L))

Substituting for Δφ:

(2π/3) = (2π/λ) * (d * (y / L))

Canceling 2π from both sides gives:

1/3 = (d * (y / L)) / λ

Rearranging for y, we find:

y = (λ/ d) * (L/3)

Plugging in the Values

Now substituting the known values:

y = (600 x 10^-9 m / 0.0025 m) * (1.2 m / 3)

Calculating this gives:

y = (240 x 10^-6 m) * (0.4) = 96 x 10^-6 m = 0.000096 m = 0.096 cm

Final Result

The distance from the central maximum where the intensity is 75.0% of the maximum is approximately 0.096 cm.