Fringe width in YDSE

In Young’s double-slit experiment, the fringe width is $\beta = \frac{\lambda D}{d}$ where $\lambda$ is the wavelength, $D$ is the screen distance, and $d$ is slit separation.

Given: $\lambda_1 = 400\,\text{nm}, \qquad \lambda_2 = 600\,\text{nm}$

Hence, $\beta_1 = \frac{\lambda_1 D}{d}, \qquad \beta_2 = \frac{\lambda_2 D}{d}$

Since $\lambda_2 > \lambda_1$, we get $\beta_2 > \beta_1$

So, Option A is correct.

Number of fringes within a fixed distance $y$

The number of fringes in a given distance is inversely proportional to fringe width: $m \propto \frac{y}{\beta}$

Therefore, $m_1 = \frac{y}{\beta_1}, \qquad m_2 = \frac{y}{\beta_2}$

Since $\beta_1 < \beta_2$, it follows that $m_1 > m_2$

So, Option B is correct.

Overlap condition for maxima and minima

• Position of the $n$th maximum: $y_n^{(\max)} = n\beta$
• Position of the $k$th minimum: $y_k^{(\min)} = \frac{2k-1}{2}\beta$

For $\lambda_2 = 600\,\text{nm}$, the 3rd maximum is at $y = 3\beta_2$

For $\lambda_1 = 400\,\text{nm}$, the 5th minimum is at $y = \frac{2(5)-1}{2}\beta_1 = \frac{9}{2}\beta_1$

Now compare: $3\beta_2 = 3\cdot \frac{600D}{d} = \frac{1800D}{d}$ $\frac{9}{2}\beta_1 = \frac{9}{2}\cdot \frac{400D}{d} = \frac{1800D}{d}$

These are equal, so the positions overlap.

Hence, Option C is correct.

Angular fringe separation

Angular fringe separation is $\Delta \theta = \frac{\lambda}{d}$

Since $\lambda_2 > \lambda_1$, $\Delta \theta_2 > \Delta \theta_1$

Therefore the angular separation for $\lambda_1$ is not greater than that for $\lambda_2$.

So, Option D is incorrect.

Final derived answer

Correct options are: $\boxed{A,\ B,\ C}$

Comparison with stored correct answer

Stored correct answer: A, B, C

My derived answer matches the stored answer.