1. In a YDSE apparatus, how many maxima can be obtained on the screen if the wavelength of light used is 200nm and d=700nm.

2. Two coherent light sources each of wavelength lambda are separated by a distance 3lambda. The maximum number of minima formed on the line AB which runs from -infinity to infinity.

To tackle your questions, we need to delve into the principles of Young's Double Slit Experiment (YDSE) and the interference of coherent light sources. Let's break down each part step by step.

Maxima in Young's Double Slit Experiment

In a YDSE setup, the number of maxima (bright fringes) that can be observed on the screen is determined by the relationship between the wavelength of light used and the distance between the slits. The formula for the position of the maxima is given by:

• y_m = (m * λ * L) / d

Where:

• y_m = position of the m-th maximum
• m = order of the maximum (0, ±1, ±2, ...)
• λ = wavelength of light
• L = distance from the slits to the screen
• d = distance between the slits

For maxima to be formed, the condition is:

• m * λ ≤ d

Given:

• Wavelength (λ) = 200 nm = 200 x 10^-9 m
• Distance between slits (d) = 700 nm = 700 x 10^-9 m

Now, we can find the maximum order of m:

• m * (200 x 10^-9) ≤ (700 x 10^-9)
• m ≤ 700 / 200 = 3.5

This means the maximum integer value for m is 3. Therefore, the possible values of m are 0, ±1, ±2, ±3. Counting these gives us:

• 1 (for m=0) + 3 (for m=±1, ±2, ±3) = 7 maxima

Thus, there will be a total of 7 maxima observed on the screen.

Minima from Two Coherent Light Sources

Now, let's consider the second part of your question regarding two coherent light sources separated by a distance of 3λ. The condition for minima in an interference pattern is given by:

• d sin(θ) = (m + 0.5)λ

Where:

• d = distance between the two sources
• m = order of the minima (0, ±1, ±2, ...)

In this case, the distance d = 3λ. Therefore, substituting this into the minima condition gives:

• 3λ sin(θ) = (m + 0.5)λ

Dividing through by λ:

• 3 sin(θ) = m + 0.5

Rearranging gives:

• m = 3 sin(θ) - 0.5

To find the maximum number of minima, we need to consider the range of sin(θ), which varies from -1 to 1. Thus:

• For sin(θ) = 1: m = 3(1) - 0.5 = 2.5
• For sin(θ) = -1: m = 3(-1) - 0.5 = -3.5

The integer values of m that can be obtained range from -3 to 2, which gives us:

• m = -3, -2, -1, 0, 1, 2

This results in a total of 6 minima (counting from -3 to 2). Therefore, the maximum number of minima formed on the line AB is 6.

In summary, for the YDSE with a wavelength of 200 nm and slit separation of 700 nm, you can observe 7 maxima. For the two coherent sources separated by 3λ, you can observe a maximum of 6 minima along the line AB.