B

Step 1: Determine the paths of each wavelength to find the position of the first maximum. The position of the maxima in a double slit interference pattern is given by:
$$ y = \frac{m \lambda L}{d} $$
where:

• $y$ = distance from the central maximum to the m-th maximum
• $m$ = order of the maximum (1 for the first maximum)
• $\lambda$ = wavelength of the light
• $L$ = distance from the slits to the screen (150 cm = 1.5 m)
• $d$ = separation between the slits (0.20 mm = 0.0002 m)

Step 2: Calculate for the wavelength $\lambda_1 = 580 \text{ nm} = 580 \times 10^{-9} \text{ m}$:
$$ y_1 = \frac{1 \times (580 \times 10^{-9}) \times 1.5}{0.0002} $$
$$ y_1 = \frac{8.7 \times 10^{-7}}{0.0002} = 0.00435 \text{ m} = 4.35 \text{ mm} $$

Step 3: Calculate for the wavelength $\lambda_2 = 700 \text{ nm} = 700 \times 10^{-9} \text{ m}$:
$$ y_2 = \frac{1 \times (700 \times 10^{-9}) \times 1.5}{0.0002} $$
$$ y_2 = \frac{1.05 \times 10^{-6}}{0.0002} = 0.00525 \text{ m} = 5.25 \text{ mm} $$

Step 4: Calculate the linear separation between the first maxima for the two wavelengths:
$$ \Delta y = y_2 - y_1 = 5.25 \text{ mm} - 4.35 \text{ mm} = 0.90 \text{ mm} $$

Step 5: Therefore, the linear separation between the first maximum (next to the central maximum) corresponding to the wavelengths 580 nm and 700 nm is 0.90 mm.