Question

In Young's experiment, using red light ($\lambda = 6600$ \AA), 60 fringes are seen in the field of view. How many fringes will be seen by using violet light ($\lambda = 4400$ A)?

• A. 10
• B. 20
• C. 45
• D. 90

Hint:

Number of fringes \(\propto \dfrac{1}{\lambda}\). A smaller wavelength (violet) gives more fringes in the same field of view.

Answer 1

The Correct Option is D

In Young's Double Slit Experiment, the fringe width is determined by the formula:

\(\beta = \dfrac{\lambda D}{d}\)

where:

• \(\beta\) is the fringe width.
• \(\lambda\) is the wavelength of the light used.
• \(D\) is the distance from the slits to the screen.
• \(d\) is the distance between the slits.

The number of fringes, \(N\), observable in a fixed field of view is inversely proportional to the fringe width \(\beta\).

Thus,

\(N \propto \dfrac{1}{\beta} \propto \dfrac{d}{\lambda D}\)

Given that 60 fringes are observed using red light with a wavelength \(\lambda_1 = 6600 \, \text{\AA}\), we can express this as:

\(N_1 = k \cdot \dfrac{d}{\lambda_1 D}\)

Similarly, for violet light with a wavelength \(\lambda_2 = 4400 \, \text{\AA}\), the number of fringes \(N_2\) is given by:

\(N_2 = k \cdot \dfrac{d}{\lambda_2 D}\)

Taking the ratio of \(N_1\) and \(N_2\):

\(\dfrac{N_2}{N_1} = \dfrac{\lambda_1}{\lambda_2}\)

Substituting the given values:

\(\dfrac{N_2}{60} = \dfrac{6600}{4400}\)

Simplifying gives:

\(\dfrac{N_2}{60} = \dfrac{3}{2}\)

Therefore:

\(N_2 = 60 \times \dfrac{3}{2} = 90\)

Hence, the number of fringes observable using violet light is: 90.