Explanation (Complete Step-by-Step)

In Young's double slit experiment, fringe shift due to insertion of a thin sheet is:

\[ \text{Fringe shift} = \frac{D}{d} (\mu - 1)t \]

Given:

\[ D = 1 \text{ m} \]

\[ d = 0.4 \text{ mm} = 0.4 \times 10^{-3} \text{ m} \]

\[ t = 20 \text{ μm} = 20 \times 10^{-6} \text{ m} \]

\[ \text{Fringe shift} = 20 \text{ mm} = 20 \times 10^{-3} \text{ m} \]

Substitute in formula:

\[ 20 \times 10^{-3} = \frac{1}{0.4 \times 10^{-3}} (\mu - 1)(20 \times 10^{-6}) \]

\[ 20 \times 10^{-3} = \frac{1}{0.4} \times 10^{3} \times (\mu - 1)(20 \times 10^{-6}) \]

\[ = 2500 (\mu - 1)(20 \times 10^{-6}) \]

\[ 20 \times 10^{-3} = 2500 \times 20 \times 10^{-6} (\mu - 1) \]

\[ 20 \times 10^{-3} = 50000 \times 10^{-6} (\mu - 1) \]

\[ 20 \times 10^{-3} = 0.05 (\mu - 1) \]

\[ \mu - 1 = \frac{20 \times 10^{-3}}{0.05} \]

\[ \mu - 1 = 0.4 \]

\[ \mu = 1.4 \]

Given \( \mu = \alpha/10 \)

\[ 1.4 = \frac{\alpha}{10} \]

\[ \alpha = 14 \]

Therefore, \( \alpha = 14 \)

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