Question

Which nature of light is supported by Young's double slit experiment? The light used in Young's double slit experiment contains two wavelengths, 6000 \AA\ and 5000 \AA. Separation between the slits is \(10^{-3}\) m and the screen is at a distance of 1 m from the slits. Find minimum distance from the central maxima where bright fringes for both the wavelengths are coincident.

Hint:

The interference pattern for two wavelengths will overlap at points where the fringe separations for both wavelengths are the same. This can be found by using the condition \(n \lambda_1 = m \lambda_2\), where \(n\) and \(m\) are integers.

Answer 1

Young's double slit experiment provides strong evidence for the wave nature of light. It demonstrates the phenomenon of interference, which occurs when light waves from two slits overlap and produce constructive or destructive interference. The experiment shows that light has a wave-like nature, as interference patterns are observed.
Now, for two different wavelengths, the condition for constructive interference (bright fringes) for each wavelength is given by the equation:
\[ \Delta y = \frac{n \lambda D}{d} \] Where:
- \(\Delta y\) is the fringe separation (distance between adjacent bright fringes),
- \(n\) is the fringe number,
- \(\lambda\) is the wavelength of the light,
- \(D\) is the distance between the screen and the slits,
- \(d\) is the distance between the slits.
For two wavelengths, the bright fringes will coincide when the fringe separation for both wavelengths is equal, i.e., when:
\[ \frac{\lambda_1 D}{d} = \frac{\lambda_2 D}{d} \] Thus, we need to find the minimum distance \(x\) where the first bright fringe for both wavelengths coincide. Let's calculate it.
We are given:
- \(\lambda_1 = 6000~\text{\AA} = 6000 \times 10^{-10}~\text{m}\),
- \(\lambda_2 = 5000~\text{\AA} = 5000 \times 10^{-10}~\text{m}\),
- \(d = 10^{-3}~\text{m}\),
- \(D = 1~\text{m}\).
To find the minimum distance, we equate the fringe separations for both wavelengths:
\[ \frac{n \lambda_1}{d} = \frac{m \lambda_2}{d} \] Simplifying, we find:
\[ n \lambda_1 = m \lambda_2 \] Substituting values:
\[ n \times 6000 \times 10^{-10} = m \times 5000 \times 10^{-10} \] Solving for \(n/m\):
\[ \frac{n}{m} = \frac{5000}{6000} = \frac{5}{6} \] Thus, the minimum value of \(n\) and \(m\) that satisfies the condition is when \(n = 5\) and \(m = 6\). Therefore, the minimum distance from the central maximum where the bright fringes for both wavelengths coincide is given by:
\[ \Delta y = \frac{5 \times 6000 \times 10^{-10} \times 1}{10^{-3}} = 3 \times 10^{-2}~\text{m} = 3~\text{cm} \] Thus, the minimum distance is 3 cm.