1. Fringe position in YDSE

For a wavelength $\lambda$, the position of the $n$-th bright fringe is

$y = n\frac{D\lambda}{d}$

where:

• slit separation $d = 0.5\,\text{mm} = 5\times 10^{-4}\,\text{m}$
• screen distance $D = 150\,\text{cm} = 1.5\,\text{m}$
• wavelengths: $\lambda_1 = 650\,\text{nm}, \qquad \lambda_2 = 520\,\text{nm}$

2. Condition for coincidence of bright fringes

Bright fringes for both wavelengths coincide when

$m\lambda_1 = n\lambda_2$

So,

$m(650) = n(520)$

$\frac{m}{n} = \frac{520}{650} = \frac{4}{5}$

Thus the smallest integers are

$m = 4, \qquad n = 5$

So the first common bright fringe from the center occurs at

$y = 4\frac{D\lambda_1}{d} = 5\frac{D\lambda_2}{d}$

3. Calculate the distance

Using $\lambda_1 = 650\,\text{nm}$:

$y = 4\cdot \frac{1.5\times 650\times 10^{-9}}{5\times 10^{-4}}$

First,

$\frac{1.5\times 650\times 10^{-9}}{5\times 10^{-4}} = 1.95\times 10^{-3}\,\text{m} = 1.95\,\text{mm}$

Therefore,

$y = 4\times 1.95\,\text{mm} = 7.8\,\text{mm}$

4. Match with options

$\boxed{7.8\,\text{mm}}$

So the correct option is C.

5. Comparison with stored answer

Stored correct answer: C

This matches our derived answer.