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: Fringe Position Formula:
Position of the $n$th secondary maximum: $y_n=\dfrac{(2n+1)\lambda D}{2a}$.

Step 2: Second Maximum:
$y_2=\dfrac{5\lambda D}{2a}$, so $\lambda=\dfrac{2ay_2}{5D}$.

Step 3: Numbers:
$a=6.6\times10^{-4}$ m, $y_2=2.8\times10^{-3}$ m, $D=1.4$ m. $\lambda=\dfrac{2\times6.6\times10^{-4}\times2.8\times10^{-3}}{7}=5.28\times10^{-7}$ m $=5280$ Å. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } 5280\ \text{\AA}} \]