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In ydse wavelength of light used is 600nm. At a point intensity is 50%of maximum value. Whatvis path difference ?
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Introduction:
In Young's double-slit experiment, light waves interfere with each other to produce a pattern of bright and dark fringes. The distance between two adjacent bright fringes is known as the path difference.

Given:
Wavelength of light used = 600nm
Intensity at a point = 50% of maximum value

Solution:
To find the path difference, we will use the formula:
path difference = (m + 1/2) * λ

where m is the order of the bright fringe.

Calculating the order of the bright fringe:
As the intensity at the point is 50% of the maximum value, it corresponds to the first bright fringe.

Calculating the path difference:
Using the formula, we get:
path difference = (1 + 1/2) * 600nm
= 900nm

Therefore, the path difference is 900nm.

Explanation:
Young's double-slit experiment is based on the interference of light waves. When a beam of light passes through two narrow slits, it diffracts and creates two coherent sources of light waves. These waves interfere with each other to produce a pattern of bright and dark fringes on a screen placed behind the slits.

The distance between two adjacent bright fringes is known as the path difference. It can be calculated using the formula (m + 1/2) * λ, where m is the order of the bright fringe and λ is the wavelength of light used.

In this case, the intensity at the point corresponds to the first bright fringe. Therefore, the order of the bright fringe is 1. Using the formula, we can calculate the path difference to be 900nm.

Conclusion:
In Young's double-slit experiment, the path difference plays an important role in determining the interference pattern. By calculating the path difference, we can determine the order of the bright fringes and predict the pattern of interference fringes.
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In ydse wavelength of light used is 600nm. At a point intensity is 50%of maximum value. Whatvis path difference ?
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