Question

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is $5\ \text{mm}$. The screen on which the diffraction pattern is displayed is at a distance of $80\ \text{cm}$ from the slit. The wavelength is $6000\ \text{\AA}$. The slit width is about

• A. $0.576\ \text{mm}$
• B. $0.348\ \text{mm}$
• C. $0.192\ \text{mm}$
• D. $0.096\ \text{mm}$

Hint:

The distance between the first minimum on the left and the first minimum on the right is simply the full angular width of the central maximum projected onto the screen ($2\lambda D / a$). Don't confuse it with the standard fringe width of double-slit interference patterns!

Answer 1

The Correct Option is C

Step 1: Understand the measurement.
In single slit diffraction, the gap between the first minimum on the left and the first minimum on the right is the full width of the central bright band. Here it is $5$ mm.

Step 2: Recall the minimum position.
The $n$-th minimum sits at $y_n=\frac{n\lambda D}{a}$, where $a$ is the slit width and $D$ is the slit-to-screen distance.

Step 3: Write the central width.
The two first minima (left and right) are at $\pm y_1$, so their separation is \[ 2y_1=\frac{2\lambda D}{a}. \]

Step 4: List the values in SI units.
$2y_1=5\times10^{-3}$ m, $D=0.8$ m, $\lambda=6000\ \text{\AA}=6\times10^{-7}$ m.

Step 5: Rearrange for the slit width.
\[ a=\frac{2\lambda D}{2y_1}=\frac{2\times6\times10^{-7}\times0.8}{5\times10^{-3}}. \]

Step 6: Do the arithmetic.
Top: $2\times6\times10^{-7}\times0.8=9.6\times10^{-7}$. Then \[ a=\frac{9.6\times10^{-7}}{5\times10^{-3}}=1.92\times10^{-4}\ \text{m}. \]

Step 7: Convert to millimetres.
$1.92\times10^{-4}\ \text{m}=0.192\ \text{mm}$, which is option (3).
\[ \boxed{0.192\ \text{mm}} \]