Animated Solution for Physics - Optics: Visible light of wavelength $6000\times 10^{-8}\text{ cm}$ falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $60^{\circ }$ from the central maximum. If the first minimum is produced at ${\theta }_1$, then ${\theta }_1$ is close to

Imagine a beam of light passing through a narrow slit. Instead of just creating a sharp shadow, the light spreads out, creating a beautiful pattern of alternating bright and dark bands on a screen. This phenomenon is known as diffraction. In this problem, we are dealing with a classic single-slit diffraction setup, and our goal is to find the angular position of the first dark band (minimum) given the position of the second one.

The key to unlocking any single-slit diffraction problem is the condition for the formation of minima (dark fringes). The angular position ${\theta }_n$ of the $n^{\text{th}}$ minimum is governed by the equation:

Here, $D$ is the width of the slit, $\lambda$ is the wavelength of the incident light, and $n$ is an integer ($1,2,3,\dots$) representing the order of the minimum.

The problem provides us with a crucial piece of information: the second diffraction minimum ($n=2$) occurs at an angle of $60^{\circ }$. Let's plug these values into our master equation:

From this, we can isolate the ratio of the wavelength to the slit width, $\frac{\lambda }{D}$. This ratio is a constant for the given setup and will be the stepping stone to our final answer.

We know that $\sin 60^{\circ }=\frac{\sqrt{3}}{2}\approx 0.866$. Substituting this value, we get:

Now, we need to find the angular position of the first minimum, which we'll call ${\theta }_1$. For the first minimum, $n=1$. Applying our master equation again:

We already calculated the value of $\frac{\lambda }{D}$ in the previous step! Let's substitute it here:

To find ${\theta }_1$, we need to determine which angle has a sine value close to $0.433$. Looking at the standard trigonometric values, we know $\sin 30^{\circ }=0.5$. Since $0.433$ is less than $0.5$, the angle must be less than $30^{\circ }$.

If we look at the given options, $25^{\circ }$ is a very strong candidate. In fact, $\sin 25^{\circ }\approx 0.422$, which is remarkably close to our calculated value of $0.433$. Therefore, we can confidently conclude that:

Did you notice something interesting? The problem explicitly gave us the wavelength of the light ($\lambda =6000\times 10^{-8}\text{ cm}$). However, we arrived at the correct answer without ever plugging this number into our calculations!

This is a classic trap in competitive exams like JEE. Examiners often provide redundant information to test your conceptual clarity and see if you can identify the most efficient path to the solution. By working with ratios instead of absolute values, we saved time and avoided messy calculations. Always simplify your equations algebraically before reaching for the numbers!