Problem 54 - Master Resource Book in JEE Main Physics

Question

In a two slits experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by \(5 \times 10^{-2} \mathrm{~m}\) towards the slits, the change in fringe width is \(3 \times 10^{-5} \mathrm{~m}\). If separation between the slits is \(10^{-3} \mathrm{~m}\), the wavelength of light used is (a) \(4500 \mathrm{~A}\) (b) \(3000 \mathrm{~A}\) (c) \(5000 \mathrm{~A}\) (d) \(6000 \AA\)

Step-by-Step Solution

Answer

The wavelength of light used is 6000 Å (option d).

Step 1: Understanding fringe width

In a two-slit experiment, the fringe width (\(\beta\)) is given by the formula \(\beta = \frac{\lambda D}{d}\), where \(\lambda\) is the wavelength of light, \(D\) is the distance from the slits to the screen, and \(d\) is the separation between the slits.

Step 2: Calculate initial fringe width change

Let the initial distance from the slits to the screen be \(D_1\) and the new distance be \(D_2 = D_1 - 5 \times 10^{-2} \mathrm{~m}\). The change in fringe width is \(3 \times 10^{-5} \mathrm{~m}\). We have \(\beta_2 - \beta_1 = 3 \times 10^{-5} \mathrm{~m}\).

Step 3: Express fringe widths using formula

The initial fringe width is \(\beta_1 = \frac{\lambda D_1}{d}\) and the new one is \(\beta_2 = \frac{\lambda (D_1 - 5 \times 10^{-2})}{d}\). The change in fringe width is \(\beta_2 - \beta_1 = \frac{\lambda (D_1 - 5 \times 10^{-2})}{d} - \frac{\lambda D_1}{d}\).

Step 4: Simplify change in fringe width equation

Simplifying the expression for change in fringe width: \[ \beta_2 - \beta_1 = \frac{\lambda}{d} (D_1 - 5 \times 10^{-2} - D_1) = \frac{-\lambda \times 5 \times 10^{-2}}{d} \]So, we know \(\frac{-\lambda \times 5 \times 10^{-2}}{d} = 3 \times 10^{-5}\).

Step 5: Solve for wavelength

Substitute \(d = 10^{-3} \mathrm{~m}\) into the equation: \[ \frac{-\lambda \times 5 \times 10^{-2}}{10^{-3}} = -3 \times 10^{-5} \]\[ \lambda \times 5 \times 10^{-2} = 3 \times 10^{-8} \]\[ \lambda = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} = 6 \times 10^{-7} \mathrm{~m} \]Convert to Angstroms: \(6 \times 10^{-7} \mathrm{~m} = 6000 \mathrm{~A}\).

Step 6: Identify the correct wavelength from options

The calculated wavelength is \(6000 \mathrm{~A}\). Reviewing the given options, (d) \(6000 \mathrm{~A}\) matches our calculation.

Key Concepts

Understanding Fringe Width

In Young's Double Slit Experiment, fringe width is how wide each bright or dark band appears on the screen. This is a key component when studying interference patterns created by light passing through two slits.
Fringe width, denoted by \( \beta \), is calculated with the formula:

• \( \beta = \frac{\lambda D}{d} \)

Here:

• \( \lambda \) is the wavelength of the light used.
• \( D \) is the distance between the slits and the screen.
• \( d \) is the separation between the two slits.

This formula tells us that the fringe width directly depends on the wavelength of the light and the distance from the slits to the screen. If either the distance \( D \) or the wavelength \( \lambda \) increases, the fringe width will increase.
Conversely, if the distance to the screen decreases, such as when moving the screen closer to the slits, the fringe width decreases. This is precisely what happens in the given exercise problem.

Monochromatic Light Explained

Monochromatic light, as used in Young's Double Slit Experiment, is light that consists of only one wavelength or color. This is crucial for creating clear, stable interference patterns.
By using light from a single wavelength, well-defined light and dark fringes are produced. These are due to the constructive and destructive interference of the light waves passing through the slits.

• Constructive interference occurs when two light waves meet in phase (their peaks align), resulting in a bright fringe.
• Destructive interference happens when they meet out of phase (peaks align with troughs), giving a dark fringe.

Without monochromatic light, the different wavelengths would overlap, leading to a blurred pattern. This makes accurate measurement of fringe width, and thus wavelength calculation, very challenging. Hence, monochromatic light ensures precision in such experiments.

Wavelength Calculation in the Experiment

Calculating the wavelength of light in Young's Double Slit Experiment involves the measurable shift in fringe width when a parameter like screen distance changes. In the given exercise, moving the screen closer affected the fringe width by a known amount.
To find the wavelength \( \lambda \), we use the formula for fringe width:

• Initial width: \( \beta_1 = \frac{\lambda D_1}{d} \)
• New width: \( \beta_2 = \frac{\lambda (D_1-5 \times 10^{-2})}{d} \)

The change is expressed as:\[\beta_2 - \beta_1 = \frac{-\lambda \times (5 \times 10^{-2})}{d}\]With a known separation of the slits \( d = 10^{-3} \mathrm{~m} \), and a change in fringe width \( 3 \times 10^{-5} \mathrm{~m} \), we calculate:

• \( \lambda \times (5 \times 10^{-2}) = 3 \times 10^{-8} \)
• Solving gives \( \lambda = 6 \times 10^{-7} \mathrm{~m} \)
• In Angstroms: \( \lambda = 6000 \mathrm{~A} \)

This shows how sensitive the method is for determining the wavelength of light, reinforcing the importance of precise measurement and consistent conditions in experiments.