A double slit apparatus is fitted with a mica strip and a...

The path difference introduced by the two strips at the center of the screen is given by:

$\Delta x_0 = (\mu_1 - 1)t - (\mu_2 - 1)t = (\mu_1 - \mu_2)t$

Substituting the given values:

$\Delta x_0 = (1.58 - 1.55) \times 0.50 \times 10^{-3} = 0.03 \times 0.50 \times 10^{-3} = 1.5 \times 10^{-5} \text{ m}$

The number of fringes by which the central maximum is shifted is:

$n_0 = \dfrac{\Delta x_0}{\lambda} = \dfrac{1.5 \times 10^{-5}}{590 \times 10^{-9}} = \dfrac{15000}{590} \approx 25.42$

The path difference at a distance $y$ from the center of the screen is:

$\Delta x = \Delta x_0 \pm \dfrac{y d}{D}$

For a maximum, the path difference must be an integral multiple of the wavelength, $\Delta x = n \lambda$. Thus, the position of the $n$-th maximum is:

$y = \dfrac{D}{d} | \Delta x_0 - n \lambda |$

To find the first maximum on either side of the center ($y = 0$), we must choose integers $n$ that are closest to $25.42$. These integers are $n = 25$ and $n = 26$.

For $n = 25$ (one side of the center):

$y_1 = \dfrac{D}{d} ( \Delta x_0 - 25 \lambda )$

$y_1 = \dfrac{1}{1.2 \times 10^{-3}} ( 1.5 \times 10^{-5} - 25 \times 590 \times 10^{-9} )$

$y_1 = \dfrac{1}{1.2 \times 10^{-3}} ( 15000 \times 10^{-9} - 14750 \times 10^{-9} )$

$y_1 = \dfrac{250 \times 10^{-9}}{1.2 \times 10^{-3}} = 2.083 \times 10^{-4} \text{ m} = 0.0208 \text{ cm} \approx 0.021 \text{ cm}$

For $n = 26$ (the other side of the center):

$y_2 = \dfrac{D}{d} ( 26 \lambda - \Delta x_0 )$

$y_2 = \dfrac{1}{1.2 \times 10^{-3}} ( 26 \times 590 \times 10^{-9} - 1.5 \times 10^{-5} )$

$y_2 = \dfrac{1}{1.2 \times 10^{-3}} ( 15340 \times 10^{-9} - 15000 \times 10^{-9} )$

$y_2 = \dfrac{340 \times 10^{-9}}{1.2 \times 10^{-3}} = 2.833 \times 10^{-4} \text{ m} = 0.0283 \text{ cm} \approx 0.028 \text{ cm}$

The first maxima are located at $0.021 \text{ cm}$ and $0.028 \text{ cm}$ from the center.

Answer: $0.021 \text{ cm}$ on one side and $0.028 \text{ cm}$ on the other side