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The fringe width $\beta$ in Young's double-slit experiment is given by:

Given values in standard SI units:

• Wavelength $\lambda = 600\,\text{nm} = 600 \times 10^{-9}\,\text{m} = 6 \times 10^{-7}\,\text{m}$
• Screen distance $D = 1.2\,\text{m}$
• Slit separation $d = 1.5\,\text{mm} = 1.5 \times 10^{-3}\,\text{m}$
  Substituting:
  $$\beta = \frac{(6 \times 10^{-7}\,\text{m})(1.2\,\text{m})}{1.5 \times 10^{-3}\,\text{m}} = \frac{7.2 \times 10^{-7}}{1.5 \times 10^{-3}} = 4.8 \times 10^{-4}\,\text{m} = 0.48\,\text{mm}$$

• Let the transmitted intensity through the first polarizer $P_1$ be $I_1$.
• If the middle polaroid sheet is oriented at an angle $\theta$ with respect to $P_1$, by Malus's Law, the transmitted intensity through it is:
  $$I_2 = I_1 \cos^2\theta$$
• Since $P_1$ and $P_2$ are crossed (at $90^\circ$), the angle between the middle sheet and $P_2$ is $(90^\circ - \theta)$. Applying Malus's Law again:
  $$I_3 = I_2 \cos^2(90^\circ - \theta) = I_1 \cos^2\theta \sin^2\theta = \frac{I_1}{4} (2 \sin\theta \cos\theta)^2 = \frac{I_1}{4} \sin^2(2\theta)$$
• $I_3$ is maximum when $\sin^2(2\theta) = 1 \implies 2\theta = 90^\circ = \pi/2 \implies \theta = 45^\circ = \pi/4$.

In Young's Double Slit Experiment, the fringe separation (fringe width) $\beta$ is given by:

Given values in standard SI units:

• Slit separation $d = 0.1\,\text{mm} = 10^{-4}\,\text{m}$
• Screen distance $D = 2\,\text{m}$
• Wavelength $\lambda = 500\,\text{nm} = 500 \times 10^{-9}\,\text{m} = 5 \times 10^{-7}\,\text{m}$
  Substituting:
  $$\beta = \frac{(5 \times 10^{-7}\,\text{m})(2\,\text{m})}{10^{-4}\,\text{m}} = \frac{10^{-6}}{10^{-4}} = 10^{-2}\,\text{m} = 1\,\text{cm}$$