In a Young’s double slit experiment, two slits are illuminated with a light of wavelength 800 nm. The line joining A1P is perpendicular to A1A2 as shown in the figure. If the first minimum is detected at P, the value of slits separation ‘a’ will be:The distance of screen form slits D = 5 cm

A
0.4 mm
B
0.2 mm
C
0.1 mm
D
0.5 mm

Solution:

Path Difference at P,
$\begin{array}{l} Δx = asin θ \\ = a \times \frac{a}{2 D} \end{array}$
For first minima.
$\begin{array}{l} Δx = \frac{λ}{2} \\ \Rightarrow \frac{a^{2}}{2 D} = \frac{λ}{2} \\ \Rightarrow a = \sqrt{λD} \\ = \sqrt{800 \times 10^{- 9} \times 5 \times 10^{- 2}} \\ = 2 \times 10^{- 4} m \\ = 0.2 mm \end{array}$

Solution:

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