Ncert Exercises & Solutions

Consider a two-slit interference arrangement (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of $λ$ such that the first minima on the screen falls at a distance D from the centre O.

Hint: The minima occurs when the path difference = $(2 n - 1) \frac{λ}{2}$ .

Step 1: Find the path travelled by the two rays.

From the given figure of two-slit interference arrangements, we can write;

$\begin{aligned} T_{2} P & = T_{2} O + OP = D + x \\ a n d T_{1} P & = T_{1} O - OP = D - x \\ S_{1} P & = \sqrt{{(S_{1} T_{1})}^{2} + {(P T_{1})}^{2}} = \sqrt{D^{2} + (D - x)^{2}} \\ and S_{2} P & = \sqrt{{(S_{2} T_{2})}^{2} + {(T_{2} P)}^{2}} = \sqrt{D^{2} + (D + x)^{2}} \end{aligned}$

Step 2: Find the path difference.

The minima will occur when, $S_{2} P - s_{1} P = (2 n - 1) \frac{λ}{2}$

$i . e ., {[D^{2} + {(D + x)}^{2}]}^{1 / 2} - {[D^{2} + {(D - x)}^{2}]}^{1 / 2} = \frac{λ}{2} [f o r f i r s t m i n i m a n = 1]$
$l f x = D;$

We can write;

${[D^{2} + 4 D^{2}]}^{1 / 2} - {[D^{2} + (0]}^{1 / 2} = \frac{λ}{2}$
$\Rightarrow {[5 D^{2}]}^{1 / 2} - {[D^{2}]}^{1 / 2} = \frac{λ}{2}$
$\Rightarrow \sqrt{5} D - D = \frac{λ}{2}$
$\Rightarrow D (\sqrt{5} - 1) = λ / 2 o r D = \frac{λ}{2 (\sqrt{5} - 1)}$
$P u t t i n g \sqrt{5} = 2.236$
$\Rightarrow \sqrt{5} - 1 = 2.236 - 1 = 1.236$
$D = \frac{λ}{2 (1.236)} = 0.404 λ$

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