Solution for (a): Single Slit Diffraction
Step 1 (Formula & Given Data): [0.5 Mark]
For a single slit of width $a$, the condition for the $n$-th dark fringe at a distance $y$ from the central maximum is given by:
$$y = \frac{n \lambda D}{a}$$Given data:
- $\lambda_1 = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$
- $\lambda_2 = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$
- $a = 1 \text{ mm} = 10^{-3} \text{ m}$
- $D = 1.5 \text{ m}$
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
For the dark fringes to coincide, the position $y$ must be the same for both wavelengths:
$$y = \frac{n_1 \lambda_1 D}{a} = \frac{n_2 \lambda_2 D}{a}$$Canceling common terms $D$ and $a$:
$$n_1 \lambda_1 = n_2 \lambda_2 \implies \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{600}{400} = \frac{3}{2}$$For the least distance, we take the smallest integers $n_1 = 3$ and $n_2 = 2$. Substituting $n_1 = 3$ into the formula for $y$:
$$y = \frac{3 \times (400 \times 10^{-9}) \times 1.5}{10^{-3}}$$ $$y = \frac{1800 \times 10^{-9} \times 1.5}{10^{-3}} = 1800 \times 1.5 \times 10^{-6}$$Step 3 (Final Concluded Answer): [0.5 Mark]
$$y = 2.7 \times 10^{-3} \text{ m} = 2.7 \text{ mm}$$Solution for (b): Young's Double Slit Experiment
Step 1 (Formula & Given Data): [0.5 Mark]
The position of the $n$-th bright fringe in YDSE is given by:
$$y = \frac{n \lambda D}{d}$$Given data:
- $\lambda_1 = 440 \text{ nm}$, $\lambda_2 = 660 \text{ nm}$
- $d = 0.6 \text{ mm} = 6 \times 10^{-4} \text{ m}$
- $D = 1.5 \text{ m}$
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
For bright fringes to coincide:
$$n_1 \lambda_1 = n_2 \lambda_2 \implies \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{660}{440} = \frac{3}{2}$$Taking $n_1 = 3$ (or $