Problem 8 - Master Resource Book in JEE Main Physics

Question

In Young's double slit experiment, the spacing between the slits is \(d\) and wavelength of light used is \(6000 \mathrm{~A}\). If the angular width of a fringe formed on a distance screen is \(1^{\circ}\), then value of \(d\) is (a) \(1 \mathrm{~mm}\) (b) \(0.05 \mathrm{~mm}\) (c) \(0.03 \mathrm{~mm}\) (d) \(0.01 \mathrm{~mm}\)

Step-by-Step Solution

Answer

The value of d is approximately 0.03 mm, corresponding to option (c).

Step 1: Understanding Fringe Width Formula

The formula to find the fringe width in Young's Double Slit Experiment is given by\[ \beta = \frac{\lambda D}{d} \]where \(\beta\) is the fringe width, \(\lambda\) is the wavelength, \(D\) is the distance to the screen, and \(d\) is the slit separation. The angular width \(\Delta \theta\) for small angles is approximately \(\Delta \theta = \frac{\beta}{D} \). So, we have that \(\Delta \theta \approx \frac{\lambda}{d} \). Here, \(\Delta \theta = 1^{\circ} = \frac{\pi}{180} \text{ radians}.\)

Step 2: Substitute Known Values

We need to solve for \(d\), the slit spacing. From the formula \(\Delta \theta \approx \frac{\lambda}{d} \), substitute the values:\[ \frac{\pi}{180} = \frac{6000 \times 10^{-10}}{d} \]

Step 3: Solve for d

Rearrange the equation to solve for \(d\):\[ d = \frac{6000 \times 10^{-10} \times 180}{\pi} \]Simplifying:\[ d \approx \frac{1080000 \times 10^{-10}}{3.14159} \]\[ d \approx 343623.5 \times 10^{-10} \]\[ d \approx 0.034 \text{ mm} \]

Step 4: Match with Options

The closest matching option for \(d = 0.034 \text{ mm} \) is option (c), which is \(0.03 \text{ mm}\).

Key Concepts

Fringe Width

In Young's Double Slit Experiment, fringe width is a crucial concept. It refers to the distance between two successive bright or dark fringes on the screen. This is important because it helps us understand how light behaves when it passes through two closely spaced slits.To calculate the fringe width, we use the formula:

• \( \beta = \frac{\lambda D}{d} \)

Here, \( \beta \) is the fringe width, \( \lambda \) is the wavelength of the light used, \( D \) is the distance from the slits to the screen, and \( d \) is the slit separation.The fringe width depends directly on the wavelength \( \lambda \) and the distance \( D \), and inversely on the slit separation \( d \). This means that the further the screen is from the slits or the longer the wavelength, the larger the fringe width will be. Conversely, if the slits are further apart, the fringe width becomes narrower.

Slit Separation

Slit separation, denoted by \( d \), is a key factor influencing the interference pattern seen in Young's Double Slit Experiment. It refers to the distance between the two slits through which the light passes. This distance is critical in determining the pattern of fringes you observe on the screen.As per the relationships defined in the experiment:

• The greater the slit separation, \( d \), the smaller the distance between fringes on the screen, since \( \beta = \frac{\lambda D}{d} \).
• If the slits are closer to each other, the fringes will be further apart, creating a wider pattern on the screen.

d affects not only the fringe width but also the overall visibility and sharpness of the pattern. By changing \( d \), students can observe changes in the interference pattern, illustrating the interplay between wave properties like wavelength and geometric variables like distance.

Angular Width

Angular width, symbolized by \( \Delta \theta \), in Young’s Double Slit Experiment measures the angular separation between consecutive bright or dark fringes. Unlike linear measurements such as across the screen, the angular width offers a way to understand fringe spreads irrespective of the observer’s distance from the fringe pattern.For situations involving small angles, the angular width is given by:

• \( \Delta \theta = \frac{\lambda}{d} \)

where \( \lambda \) is the wavelength and \( d \) is the slit separation. In this expression, \( \Delta \theta \) reflects how much the pattern is spread out from the center of the slits.In our specific exercise involving an angle of \(1^{\circ}\), converting degrees to radians is necessary for calculations, yielding \( \Delta \theta = \frac{\pi}{180} \). This expression highlights that as the wavelength \( \lambda \) increases or the slit separation \( d \) decreases, the angular width grows, allowing a broader spread of fringe patterns. Understanding angular width is essential for correctly predicting where fringes will appear in numerous optical experiments.