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: Understanding the Concept:
In single slit diffraction, secondary maxima occur at \(a\sin\theta = \left(n + \frac12\right)\lambda\), where \(n = 1, 2, \dots\). For the second order maximum, \(n = 2\), so \(a\sin\theta = \frac52\lambda\).

Step 2: Small angles:
\(\sin\theta \approx \frac yD\). So
\[ a\cdot\frac yD = \frac{5\lambda}{2} \Rightarrow a = \frac{5\lambda D}{2y} \]

Step 3: Substitute:
\(\lambda = 580\times10^{-9}\) m, \(D = 2.5\) m, \(y = 14.5\times10^{-3}\) m.
\[ a = \frac{5\times580\times10^{-9}\times2.5}{2\times14.5\times10^{-3}} = \frac{7.25\times10^{-6}}{29\times10^{-3}} = 2.5\times10^{-4}\text{ m} \]

Step 4: Result:
\(a = 0.25\times10^{-3}\) m.

Step 5: Why the other options are wrong.
\(0.12\times10^{-3}\) m would follow from using \(a\sin\theta = 5\lambda\)-type slips, and \(0.5\times10^{-3}\) m from using \(2\lambda\) instead of \(\frac52\lambda\).

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