In Young's double slit experiment, we investigate how changes in certain parameters affect the interference pattern produced. To answer this question, let's analyze each option with respect to the changes mentioned: changing the wavelength of light from 7000 Å to 3500 Å and doubling the separation between the slits.
The formula for fringe width in Young's double slit experiment is given by:
\(\beta = \frac{\lambda \cdot D}{d}\)
where:
- \(\beta\) is the fringe width (distance between consecutive bright or dark fringes).
- \(\lambda\) is the wavelength of light used.
- \(D\) is the distance between the slits and the screen.
- \(d\) is the separation between the slits.
Given the changes:
- The wavelength changes from 7000 Å to 3500 Å.
- The slit separation \(d\) is doubled.
Substituting these changes into the fringe width formula:
\(\beta' = \frac{3500 \cdot D}{2d} = \frac{1}{2} \times \frac{7000 \cdot D}{d} = \frac{1}{2} \beta\)
This calculation shows that the fringe width is halved. Now, let's assess each given statement:
- The width of the fringes changes:
This is true as shown by the calculation; the fringe width is halved. - The color of bright fringes changes:
This is true because the color is determined by the wavelength of light, which has also changed. - The separation between successive bright fringes changes:
This is true, again confirmed by the calculation showing that fringe width (separation between consecutive bright fringes) is halved. - The separation between successive dark fringes remains unchanged:
This is not true. Since the formula for fringe width applies to both bright and dark fringes, a reduction in fringe width also implies a change in separation between successive dark fringes. Thus, the separation between dark fringes is also halved.
Hence, the correct answer is that "The separation between successive dark fringes remains unchanged" is not true for this situation.