Sodium vapour light of wave length 5893A° is used first in a fresnels biprism set up .a total of 60 fringes are observed in the field of view of the eye piece .determine the number of fringes that would be observed in the same field of view. If the source is replaced with another source of wave length 5300A° ?

To determine the number of fringes observed when the wavelength of the light source changes, we can use the relationship between the wavelength of light and the number of fringes produced in an interference pattern. The number of fringes is inversely proportional to the wavelength of the light used. Let's break this down step by step.

Understanding the Relationship Between Wavelength and Fringes

In an interference setup like the Fresnel biprism, the number of fringes (N) observed can be calculated using the formula:

• N = (D * λ) / d

Where:

• N = Number of fringes
• D = Distance between the fringes
• λ = Wavelength of the light
• d = Separation between the slits (or equivalent in the biprism setup)

From this formula, we can see that if the wavelength (λ) decreases, the number of fringes (N) increases, assuming D and d remain constant. Conversely, if the wavelength increases, the number of fringes decreases.

Calculating the Number of Fringes for Different Wavelengths

Initially, with sodium vapor light of wavelength 5893 Å (angstroms), we observe 60 fringes. Now, we need to find out how many fringes would be observed if we switch to a light source with a wavelength of 5300 Å.

Since the number of fringes is inversely proportional to the wavelength, we can set up a ratio:

• N1 / N2 = λ2 / λ1

Where:

• N1 = Number of fringes with the first wavelength (60 fringes)
• N2 = Number of fringes with the second wavelength (unknown)
• λ1 = First wavelength (5893 Å)
• λ2 = Second wavelength (5300 Å)

Substituting the Values

Now, substituting the known values into the equation:

• 60 / N2 = 5300 / 5893

Cross-multiplying gives us:

• 60 * 5893 = N2 * 5300

Calculating the left side:

• 60 * 5893 = 353580

Now, we can solve for N2:

• N2 = 353580 / 5300

Calculating N2:

• N2 ≈ 66.7

Final Result

Since the number of fringes must be a whole number, we round this to 67 fringes. Therefore, if the source is replaced with another source of wavelength 5300 Å, approximately 67 fringes would be observed in the same field of view.

This example illustrates how changing the wavelength of light affects the interference pattern, demonstrating the fundamental principles of wave optics in a practical scenario.