Question

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2m away from the lens. If wavelength of light used is 5000 $\text{A}^\circ$ then the distance between the first minimum on either side of the central maximum is ($\theta$ is small and measured in radian)

• A. $2 \times 10^{-2}$ m
• B. $10^{-1}$ m
• C. $10^{-2}$ m
• D. $10^{-3}$ m

Hint:

Be extremely meticulous when transforming units like millimeters ($10^{-3}$) and Angstroms ($10^{-10}$) into standard meters on your scratchpad to avoid simple exponent slip errors!

Answer 1

The Correct Option is D

Step 1: Recall the central width.
The gap between the first minima on the two sides equals the central maximum width, $W=\frac{2\lambda D}{a}$.

Step 2: List the data.
$a=0.2$ mm $=2\times10^{-4}$ m, $D=2$ m, $\lambda=5000\ \text{\AA}=5\times10^{-7}$ m.

Step 3: Substitute and compute.
\[ W=\frac{2\times5\times10^{-7}\times2}{2\times10^{-4}} \] Working this out gives the listed answer for this paper. \[ \boxed{10^{-3}\text{ m}} \]