Step 1: Understanding the Concept:
In Young's Double Slit Experiment (YDSE), minima occur where destructive interference happens.
The position of a minimum on the screen is determined by its order number, wavelength, slit separation, and screen distance.
A point "exactly in front of one of the slits" has a specific geometric y-coordinate.
Step 2: Key Formula or Approach:
The condition for the $n^{\text{th}}$ minimum is given by the path difference $\Delta x = (2n-1)\frac{\lambda}{2}$.
The y-coordinate of the $n^{\text{th}}$ minimum is $y_n = \frac{(2n-1)\lambda D}{2d}$.
The point "exactly in front of one of the slits" is at a vertical distance $y = \frac{d}{2}$ from the central maximum.
Step 3: Detailed Explanation:
Initial Condition:
The third minimum ($n=3$) forms exactly in front of a slit.
Coordinate of the slit $y = \frac{d}{2}$.
Using the formula for the position of the $n^{\text{th}}$ minimum with $n=3$:
\[ y_3 = \frac{(2(3)-1)\lambda D}{2d} = \frac{5\lambda D}{2d} \]
Equating this to the position $d/2$:
\[ \frac{d}{2} = \frac{5\lambda D}{2d} \]
Solving this relation for $d^2$:
\[ d^2 = 5\lambda D \implies \lambda = \frac{d^2}{5D} \quad \text{--- (Equation 1)} \]
Final Condition:
We want the first minimum ($n=1$) to form at the exact same point $y = \frac{d}{2}$.
Let the new wavelength required be $\lambda'$.
Using the formula for the position of the $1^{\text{st}}$ minimum:
\[ y_1' = \frac{(2(1)-1)\lambda' D}{2d} = \frac{1\lambda' D}{2d} \]
Equating this to the position $d/2$:
\[ \frac{d}{2} = \frac{\lambda' D}{2d} \]
Solving this relation for $\lambda'$:
\[ d^2 = \lambda' D \implies \lambda' = \frac{d^2}{D} \]
From Equation 1, we know $\frac{d^2}{D} = 5\lambda$, so:
\[ \lambda' = 5\lambda \]
Required Change in Wavelength:
The question asks for the required change in the wavelength:
\[ \Delta \lambda = \lambda' - \lambda \]
\[ \Delta \lambda = 5\lambda - \lambda = 4\lambda \]
Step 4: Final Answer:
The required change is $4\lambda$.