CBSE · Class 12 · Physics · Wave OpticsIn Young's double-slit experiment, the distance between the two slits is $0.15\text{ mm}$, and the screen is kept at a distance of $1.0\text{ m}$ from the slits. The wavelength of light used is $589\text{ nm}$. Calculate: (i) the fringe width, and (ii) the distance of the 4th bright fringe from the central maximum.

Given Data:

• Distance between the slits ($d$) = $0.15\text{ mm} = 0.15 \times 10^{-3}\text{ m} = 1.5 \times 10^{-4}\text{ m}$
• Distance of the screen from the slits ($D$) = $1.0\text{ m}$
• Wavelength of light ($\lambda$) = $589\text{ nm} = 589 \times 10^{-9}\text{ m} = 5.89 \times 10^{-7}\text{ m}$
• Order of the bright fringe ($n$) = $4$

(i) Calculation of Fringe Width ($\beta$):\nThe formula for fringe width in Young's double-slit experiment is:

$$\beta = \frac{\lambda D}{d}$| \nSubstituting the given values into the formula: $$\beta = \frac{(5.89 \times 10^{-7}\text{ m}) \times (1.0\text{ m})}{1.5 \times 10^{-4}\text{ m}}$$

$$\beta = \frac{5.89 \times 10^{-7}}{1.5 \times 10^{-4}}$$

$$\beta = 3.926 \times 10^{-3}\text{ m} = 3.93\text{ mm}$$ \nThus, the fringe width is $3.93\text{ mm}$.

(ii) Calculation of the Distance of the 4th Bright Fringe ($x_n$):\nThe formula for the distance of the $n$-th bright fringe from the central maximum is:

$$x_n = n \left(\frac{\lambda D}{d}\right) = n\beta$$ \nFor the 4th bright fringe ($n = 4$): $$x_4 = 4 \times \beta$$

$$x_4 = 4 \times (3.926 \times 10^{-3}\text{ m})$$

$$x_4 = 1.5704 \times 10^{-2}\text{ m} = 15.70\text{ mm}$$ \nThus, the distance of the 4th bright fringe from the central maximum is $15.70\text{ mm}$ (or $1.57\text{ cm}$).