Step 1: Understanding the Concept:
In Young's Double Slit Experiment, the fringe width (the distance between two consecutive bright or dark fringes) depends on the wavelength of the light used.
If the experimental setup (screen size, slit distance, etc.) remains fixed, the total width of the interference pattern observed on the screen is constant.
Step 2: Key Formula or Approach:
The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the slit separation.
Let $L$ be the fixed width of the observation field on the screen. The number of fringes $n$ observed in this field is $n = \frac{L}{\beta}$.
Substituting $\beta$: $n = \frac{L d}{\lambda D}$.
Since $L, d,$ and $D$ are constant for a given setup, the product of the number of fringes and the wavelength is a constant:
\[ n_1 \lambda_1 = n_2 \lambda_2 = \text{constant} \]
Step 3: Detailed Explanation:
Given values:
Initial wavelength, $\lambda_1 = 600\text{ nm}$
Initial number of fringes, $n_1 = 18$
New wavelength, $\lambda_2 = 400\text{ nm}$
We need to find the new number of fringes, $n_2$.
Using the relation established above:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
\[ 18 \times 600\text{ nm} = n_2 \times 400\text{ nm} \]
Solving for $n_2$:
\[ n_2 = \frac{18 \times 600}{400} \]
\[ n_2 = 18 \times \frac{6}{4} = 18 \times 1.5 \]
\[ n_2 = 27 \]
Step 4: Final Answer:
The number of fringes observed will be 27.