Question

Light of wavelength \(580\) nm is incident normally on a slit of width '\(a\)'. The distance between slit and screen is \(2.5\) m and the distance of the second order maximum from the centre of the screen is \(14.5\) mm in a diffraction pattern. The value of '\(a\)' is

• A. \(0.12\times 10^{-3}\,\text{m}\)
• B. \(0.25\times 10^{-3}\,\text{m}\)
• C. \(0.36\times 10^{-3}\,\text{m}\)
• D. \(0.50\times 10^{-3}\,\text{m}\)

Hint:

Secondary maxima of a single slit lie at a sin(theta) = (n + 1/2) lambda; the second one has n = 2.

Answer 1

The Correct Option is B

Step 1: Position formula:
The second secondary maximum is at $y = \frac{(2n+1)\lambda D}{2a}$ with $n = 2$, which equals $\frac{5\lambda D}{2a}$.

Step 2: Compute the numerator:
$5\times580\ \text{nm}\times2.5\ \text{m} = 7250\ \text{nm m} = 7.25\times10^{-6}\ \text{m}^2$.

Step 3: Divide:
$a = \frac{7.25\times10^{-6}}{2\times0.0145} = \frac{7.25\times10^{-6}}{0.029} = 2.5\times10^{-4}$ m.

Final Answer:
The slit width is 0.25 mm, option (B). \[ \boxed{0.25\times10^{-3}\text{ m}} \]