In a youth`s double-slit experiment,h=500nm d=1nm andD=1m.the minimum distance from the central maximum for whi8 the intensity is half of the maximum intensity is-

In the context of the double-slit experiment, we can analyze how interference patterns are formed and how to calculate the position where the intensity drops to half the maximum intensity. Let's break this down step by step, using the information you've provided: the wavelength (λ) is 500 nm, the slit separation (d) is 1 nm, and the distance to the screen (D) is 1 m.

Understanding the Double-Slit Experiment

The double-slit experiment demonstrates the wave nature of light through interference patterns created by overlapping waves from two slits. The intensity of light at any point on the screen can be determined by the constructive and destructive interference of these waves.

Intensity in the Interference Pattern

The intensity \( I \) at a point on the screen can be expressed using the following formula:

• \( I = I_{max} \cos^2 \left( \frac{\pi d y}{\lambda D} \right) \)

Where:

• \( I_{max} \) is the maximum intensity.
• \( d \) is the distance between the slits.
• \( y \) is the distance from the central maximum to the point of interest on the screen.
• \( \lambda \) is the wavelength of the light.
• \( D \) is the distance from the slits to the screen.

Finding the Minimum Distance for Half Maximum Intensity

To find the point where the intensity is half of the maximum intensity, we set up the equation:

• \( I = \frac{1}{2} I_{max} \)

Substituting into the intensity equation gives:

• \( \frac{1}{2} I_{max} = I_{max} \cos^2 \left( \frac{\pi d y}{\lambda D} \right) \)

This simplifies to:

• \( \cos^2 \left( \frac{\pi d y}{\lambda D} \right) = \frac{1}{2} \)

Taking the square root of both sides, we have:

• \( \cos \left( \frac{\pi d y}{\lambda D} \right) = \frac{1}{\sqrt{2}} \)

Calculating the Angle

The cosine of \( \frac{\pi}{4} \) radians (or 45 degrees) is \( \frac{1}{\sqrt{2}} \). Therefore, we can equate:

• \( \frac{\pi d y}{\lambda D} = \frac{\pi}{4} \)

Now, solving for \( y \):

• \( y = \frac{\lambda D}{4d} \)

Plugging in the Values

Now, we can substitute the values into our equation. Remember:

• \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} \)
• \( d = 1 \text{ nm} = 1 \times 10^{-9} \text{ m} \)
• \( D = 1 \text{ m} \)

Substituting these values gives:

• \( y = \frac{(500 \times 10^{-9} \text{ m})(1 \text{ m})}{4(1 \times 10^{-9} \text{ m})} \)

This simplifies to:

• \( y = \frac{500 \times 10^{-9}}{4 \times 10^{-9}} = \frac{500}{4} = 125 \text{ m} \)

The Result

Thus, the minimum distance from the central maximum to the point where the intensity is half of the maximum intensity is 125 m. This result illustrates how significant the slit separation and the wavelength of light can be in determining the interference pattern in such experiments.