Fringe positions in YDSE

For a wavelength $\lambda$, the position of the $m$-th bright fringe is $y = m\beta = m\frac{D\lambda}{d}$ where:

• $D = 7\,\text{m}$
• $d = 0.35\,\text{mm} = 3.5\times 10^{-4}\,\text{m}$

Condition for coincidence of bright fringes

Let the bright fringes for wavelengths $\lambda_1 = 800\,\text{nm}$ and $\lambda_2 = 600\,\text{nm}$ coincide at some point.

Then $m_1\lambda_1 = m_2\lambda_2$ $m_1(800) = m_2(600)$ $4m_1 = 3m_2$

The smallest non-zero integers satisfying this are $m_1 = 3, \quad m_2 = 4$

Find the shortest distance from central maximum

Using either wavelength: $y = m_1\frac{D\lambda_1}{d} = 3\cdot \frac{7\cdot 800\times 10^{-9}}{3.5\times 10^{-4}}$

Simplify: $\frac{7}{3.5\times 10^{-4}} = 2\times 10^4$

So, $\beta_1 = \frac{D\lambda_1}{d} = 2\times 10^4 \times 800\times 10^{-9}$ $= 16\times 10^{-3}\,\text{m} = 16\,\text{mm}$

Therefore, $y = 3\times 16 = 48\,\text{mm}$

(Checking with $600\,\text{nm}$: fringe width $=12\,\text{mm}$, and $4\times 12 = 48\,\text{mm}$.)

Final Answer

The shortest distance from the central bright maximum where the bright fringes coincide is $\boxed{48\,\text{mm}}$