Question

In a single slit diffraction pattern, the distance between the plane of the slit and the screen is \(1.4\) m. The width of the slit is \(0.66\) mm. The second maximum is formed at the distance of \(2.8\) mm from the center of the screen. The wavelength of light used is

• A. \(6500 \text{Å}\)
• B. \(5600 \text{Å}\)
• C. \(5280 \text{Å}\)
• D. \(4600 \text{Å}\)

Hint:

Secondary maxima occur at a sin theta = (n + 1/2) lambda, so the second maximum has n = 2.

Answer 1

The Correct Option is C

Step 1: Condition for Secondary Maxima:
In a single slit pattern, the \(n\)th secondary maximum is at \(a\sin\theta=\left(n+\tfrac12\right)\lambda\). For the second maximum, \(n=2\): \(a\sin\theta=\dfrac52\lambda\).

Step 2: Small Angle:
\(\sin\theta\approx\dfrac yD\), with \(y=2.8\) mm and \(D=1.4\) m. So \(\dfrac{ay}D=\dfrac52\lambda\).

Step 3: Solve for the Wavelength:
\[ \lambda=\frac{2ay}{5D}=\frac{2\times0.66\times10^{-3}\times2.8\times10^{-3}}{5\times1.4} \]
\[ \lambda=\frac{3.696\times10^{-6}}{7}=5.28\times10^{-7}\ \text{m}=5280\ \text{\AA} \]

Step 4: Check the Other Options:
Using \(2\lambda\) or \(3\lambda/2\) instead of \(5\lambda/2\) would give 6600 Å or 8800 Å, which do not match the options. So (C) is correct.

Final Answer:
The wavelength is 5280 Å, option (C). \[ \boxed{\text{(C) } 5280\ \text{\AA}} \]