Question

A beam of coherent light of wavelength $550\text{ nm}$ is incident normal to the plane of a pair of two slits $S_1$ and $S_2$ each of width $1.2 \times 10^{-6}\text{ m}$ separated by $1.1\text{ mm}$. Dark and bright fringes are observed on a screen $2.2\text{ m}$ away from the plane of the slits.
Calculate :
(I) fringe width.
(II) distance of the second dark fringe from the central maximum.
(III) what will happen when the entire apparatus is immersed in water.

Hint:

Remember that optical path lengths change when a system is placed in a medium. For any medium of refractive index $n$, the entire interference pattern scales down by a factor of $1/n$. Thus, fringe width, fringe positions, and angular widths all decrease by exactly $1/n$.

Answer 1

Concept: This problem explores the wave theory of light using Young's Double Slit Experiment (YDSE). Interference fringes are produced when coherent light beams from two closely spaced slits overlap on a distant screen. The primary physical parameters involved are:
• Fringe Width ($\beta$): The separation distance between any two consecutive bright or consecutive dark fringes, defined mathematically as: \[ \beta = \frac{\lambda D}{d} \]
• Position of Dark Fringes ($y_n$): Destructive interference occurs when the path difference is an odd integral multiple of half-wavelengths ($ \Delta x = (2n-1)\frac{\lambda}{2} $). The position of the $n$-th dark fringe measured from the central bright maximum is given by: \[ y_n = (2n - 1)\frac{\lambda D}{2d} = \left(n - \frac{1}{2}\right)\beta \]
• Medium Refractive Index Influence: When the apparatus is completely immersed in a medium of refractive index $n_m$ (such as water, where $n_w = \frac{4}{3}$), the velocity of light changes, which modifies the operational wavelength ($\lambda' = \frac{\lambda}{n_m}$). Since fringe width is proportional to wavelength, the new fringe width scales down inversely with the refractive index ($\beta' = \frac{\beta}{n_m}$).

Step 1: Extract and convert all given data into SI units. From the problem statement, we harvest the following parameters:
• Wavelength of light, $\lambda = 550\text{ nm} = 550 \times 10^{-9}\text{ m}$
• Separation distance between the two slits, $d = 1.1\text{ mm} = 1.1 \times 10^{-3}\text{ m}$
• Distance of the observation screen from the slits, $D = 2.2\text{ m}$
• Width of individual slits, $w = 1.2 \times 10^{-6}\text{ m}$ (Note: Individual slit width limits the diffraction envelope but does not alter the double-slit interference fringe spacing calculations unless single-slit envelope restrictions are requested).

Step 2: Solve Part (I) - Calculate the fringe width ($\beta$). Using the standard interference formula for fringe width: \[ \beta = \frac{\lambda D}{d} \] Substitute the values: \[ \beta = \frac{(550 \times 10^{-9}\text{ m}) \times 2.2\text{ m}}{1.1 \times 10^{-3}\text{ m}} \] Simplify the numerical calculation by grouping the decimal constants and the powers of 10: \[ \beta = \frac{550 \times 2.2}{1.1} \times \frac{10^{-9}}{10^{-3}} \] Notice that $\frac{2.2}{1.1} = 2$. Substituting this yields: \[ \beta = 550 \times 2 \times 10^{-9 - (-3)} \] \[ \beta = 1100 \times 10^{-6}\text{ m} \] Converting this value into millimeters for a standard representation: \[ \beta = 1.1 \times 10^{-3}\text{ m} = 1.1\text{ mm} \] The fringe width of the interference pattern is exactly $1.1\text{ mm}$.

Step 3: Solve Part (II) - Calculate the distance of the second dark fringe from the central maximum. The position equation for dark fringes from the central maximum is: \[ y_n = (2n - 1)\frac{\lambda D}{2d} = (2n - 1)\frac{\beta}{2} \] For the second dark fringe, we substitute $n = 2$: \[ y_{2d} = (2(2) - 1)\frac{\beta}{2} = (4 - 1)\frac{\beta}{2} = \frac{3}{2}\beta = 1.5\beta \] We already found $\beta = 1.1\text{ mm}$ in part (I). Substitute this value into the equation: \[ y_{2d} = 1.5 \times 1.1\text{ mm} = 1.65\text{ mm} \] Alternatively, in base meters: \[ y_{2d} = 1.65 \times 10^{-3}\text{ m} \] The distance of the second dark fringe from the central maximum is $1.65\text{ mm}$.

Step 4: Solve Part (III) - Analyze immersion of the apparatus in water. When the entire experimental configuration is immersed in water, the physical distances $d$ and $D$ remain unaffected. However, the optical properties of light waves change based on the refractive index of water ($n_w \approx \frac{4}{3} = 1.33$). The wavelength of light in water ($\lambda'$) is compressed: \[ \lambda' = \frac{\lambda}{n_w} \] Since the fringe width $\beta$ is directly proportional to the wavelength ($\beta \propto \lambda$), the new fringe width $\beta'$ becomes: \[ \beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{n_w d} = \frac{\beta}{n_w} \] Substitute our parameters ($\beta = 1.1\text{ mm}$ and $n_w = \frac{4}{3}$): \[ \beta' = \frac{1.1\text{ mm}}{\frac{4}{3}} = \frac{3 \times 1.1\text{ mm}}{4} = \frac{3.3\text{ mm}}{4} = 0.825\text{ mm} \] Converting back to meters: \[ \beta' = 0.825 \times 10^{-3}\text{ m} = 8.25 \times 10^{-4}\text{ m} \] Summary of Effects:
• The wavelength reduces, causing the interference fringes to shrink closer together.
• The fringe width decreases from $1.1\text{ mm}$ to $0.825\text{ mm}$.
• The entire interference pattern becomes more closely packed or compressed towards the central bright maximum.