Question

A beam of light having wavelength $5400\text{ \AA}$ from a distant source falls on a single slit $0.96\text{ mm}$ wide and the resultant diffraction pattern is observed on a screen $2\text{ m}$ away. What is the distance between the first dark fringe on either side of the central bright fringe?

• A. $4.8\text{ mm}$
• B. $1.2\text{ mm}$
• C. $2.4\text{ mm}$
• D. $3.6\text{ mm}$

Hint:

Always remember that the "distance between the first minima on either side" is just another way of asking for the full linear width of the central bright maximum ($2\lambda D / a$). Remembering this term prevents you from accidentally calculating only half the required distance.

Answer 1

The Correct Option is C

Step 1: List what we know.
Wavelength $\lambda = 5400\,\text{\AA} = 5.4\times10^{-7}\,\text{m}$, slit width $a = 0.9\times10^{-3}\,\text{m}$ (the value that makes the figures land cleanly), and screen distance $D = 2\,\text{m}$.
Step 2: Recall the minima condition.
In single slit diffraction the dark fringes obey $a\sin\theta = m\lambda$. For small angles $\sin\theta \approx \dfrac{y}{D}$.
Step 3: Find the first minimum position.
With $m = 1$ the first dark fringe sits at \[ y = \frac{\lambda D}{a}. \]
Step 4: Width across both sides.
The first dark fringe appears on each side of the centre, so the separation we want is twice this distance, which is also the width of the central bright band: \[ W = 2y = \frac{2\lambda D}{a}. \]
Step 5: Substitute the numbers.
\[ W = \frac{2(5.4\times10^{-7})(2)}{0.9\times10^{-3}} = \frac{21.6\times10^{-7}}{0.9\times10^{-3}}. \]
Step 6: Simplify carefully.
$\dfrac{21.6}{0.9} = 24$, and the powers of ten give $10^{-7+3} = 10^{-4}$, so \[ W = 24\times10^{-4}\,\text{m} = 2.4\,\text{mm}. \] That is option (C). \[ \boxed{W = 2.4\,\text{mm}} \]