Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be:

<p>In Young's double-slit experiment, the fringe width ($\beta$) can be calculated using the formula:</p> <p>$\beta = \frac{\lambda' \cdot D}{d}$</p> <p>where:</p> <p><p>$\lambda'$ is the wavelength of light in the medium,</p></p> <p><p>$D$ is the distance between the slits and the screen,</p></p> <p><p>$d$ is the separation between the slits.</p></p> <p>The wavelength in the medium ($\lambda'$) is given by:</p> <p>$\lambda' = \frac{\lambda_0}{n}$</p> <p>where:</p> <p><p>$\lambda_0 = 690 \, \text{nm}$ is the wavelength of light in air,</p></p> <p><p>$n = 1.44$ is the refractive index of the liquid.</p></p> <p>Substituting the values,</p> <p>$\lambda' = \frac{690 \, \text{nm}}{1.44} \approx 479.17 \, \text{nm}$</p> <p>Now, convert $\lambda'$ to meters:</p> <p>$\lambda' = 479.17 \, \text{nm} = 479.17 \times 10^{-9} \, \text{m}$</p> <p>Next, use the values for $D$ and $d$:</p> <p><p>$D = 0.72 \, \text{m}$</p></p> <p><p>$d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$</p></p> <p>Calculating the fringe width:</p> <p>$$ \beta = \frac{479.17 \times 10^{-9} \, \text{m} \cdot 0.72 \, \text{m}}{1.5 \times 10^{-3} \, \text{m}} $$</p> <p>$$ \beta \approx \frac{344.2 \times 10^{-9} \, \text{m}}{1.5 \times 10^{-3} \, \text{m}} $$</p> <p>$\beta \approx 0.229 \times 10^{-3} \, \text{m} = 0.229 \, \text{mm}$</p> <p>The closest answer choice is:</p> <p>Option D: 0.23 mm</p>