1. Condition for first minimum in YDSE

For two coherent slits separated by distance $a$, the path difference at a point making angle $\theta$ with the central axis is

$\Delta = a\sin\theta$

For the first minimum,

$a\sin\theta = \frac{\lambda}{2}$

2. Geometry from the figure description

The statement says the line joining $A_1P$ is perpendicular to $A_1A_2$. This means triangle geometry is such that point $P$ lies straight ahead of slit $A_1$ on the screen, while slit $A_2$ is displaced by separation $a$.

So distances are:

• $A_1P = D = 5\text{ cm}$
• $A_2P = \sqrt{D^2 + a^2}$

Hence the path difference at $P$ is

$\Delta = A_2P - A_1P = \sqrt{D^2+a^2}-D$

Since $P$ is the first minimum,

$\sqrt{D^2+a^2}-D = \frac{\lambda}{2}$

3. Substitute values

Given:

$D = 5\text{ cm} = 5\times 10^{-2}\text{ m}$ $\lambda = 800\text{ nm} = 8\times 10^{-7}\text{ m}$

Thus,

$\sqrt{D^2+a^2} = D + \frac{\lambda}{2}$

Squaring both sides,

$D^2+a^2 = \left(D+\frac{\lambda}{2}\right)^2$

$a^2 = D\lambda + \frac{\lambda^2}{4}$

Since $\lambda^2/4$ is negligible compared to $D\lambda$,

$a^2 \approx D\lambda$

Now,

$a \approx \sqrt{(5\times 10^{-2})(8\times 10^{-7})}$

$a = \sqrt{4\times 10^{-8}}$

$a = 2\times 10^{-4}\text{ m}$

$a = 0.2\text{ mm}$

4. Option check

• A: $0.2\text{ mm}$ ✅
• B: $0.5\text{ mm}$ ❌
• C: $0.4\text{ mm}$ ❌
• D: $0.1\text{ mm}$ ❌

Therefore, the correct answer is A.