Statement-1: In Young's double slit experiment the two slits are at distance d apart. Interference pattern is observed on a screen at distance D from the slits. At a poit on the screen when it is directly opposite to one of the slits, a dard fringe is observed then the wavelength of wave is proportional of square of distance of two slits.
Statement-2: In Young's double slit experiment, for identical slits, the intensity of a dark fringe is zero.

Options

A. Statement-1 is true, statement-2 is true: Statement-2 is a correct explanation for statement-1
B. Statement-1 is true, statement-2 true, statement-2 is not a correct explanation for statement-1
C. Statement-1 is true, Statement-2 is false.
D. Statement-1 is false, Statement-2 is true.

Answer

To solve the problem, we need to analyze both statements regarding Young's double slit experiment and determine their validity and relationship.

### Step 1: Analyze Statement 1
**Statement 1:** In Young's double slit experiment, the two slits are at a distance \(d\) apart. Interference pattern is observed on a screen at distance \(D\) from the slits. At a point on the screen directly opposite to one of the slits, a dark fringe is observed, then the wavelength of the wave is proportional to the square of the distance between the two slits.

- In Young's double slit experiment, the condition for dark fringes (destructive interference) is given by:
\[ \Delta x = \frac{(m + \frac{1}{2}) \lambda}{d} \]
where \(\Delta x\) is the path difference, \(m\) is an integer (order of the dark fringe), \(\lambda\) is the wavelength, and \(d\) is the distance between the slits.

- For a point directly opposite one of the slits, the path difference can be approximated as:
\[ \Delta x = \frac{d^2}{2D} \]
(using small angle approximation).

- Setting the path difference equal to \(\frac{(m + \frac{1}{2}) \lambda}{d}\) for dark fringes, we find:
\[ \frac{d^2}{2D} = \frac{(m + \frac{1}{2}) \lambda}{d} \]
Rearranging gives:
\[ \lambda = \frac{d^3}{2D(m + \frac{1}{2})} \]
This shows that \(\lambda\) is proportional to \(d^2\) when considering the relationship in the context of the dark fringe condition.

### Conclusion for Statement 1:
Statement 1 is **true**.

### Step 2: Analyze Statement 2
**Statement 2:** In Young's double slit experiment, for identical slits, the intensity of a dark fringe is zero.

- The intensity at a dark fringe is given by the formula:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
where \(\phi\) is the phase difference. For dark fringes, the phase difference is \(\pi\) (or an odd multiple of \(\pi\)), leading to:
\[ I = I_1 + I_2 - 2\sqrt{I_1 I_2} \]
If \(I_1 = I_2\) (identical slits), then:
\[ I = I_1 + I_1 - 2\sqrt{I_1 I_1} = 0 \]
Thus, the intensity at a dark fringe is indeed zero.

### Conclusion for Statement 2:
Statement 2 is **true**.

### Final Conclusion:
Both statements are true, but Statement 2 does not explain Statement 1. Therefore, the answer is that both statements are true, but Statement 2 is not a correct explanation for Statement 1.

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