• 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 $\sin^2(2\theta) = 1 \implies 2\theta = 90^\circ = \pi/2 \implies \theta = 45^\circ = \pi/4$.

The linear width of the central maximum in single slit Fraunhofer diffraction is:

Given values in SI units:

• Linear width $\beta_0 = 5\,\text{mm} = 5 \times 10^{-3}\,\text{m}$
• Slit width $a = 0.1\,\text{mm} = 10^{-4}\,\text{m}$
• Distance to screen $D = 50\,\text{cm} = 0.5\,\text{m}$
  Rearranging for wavelength $\lambda$:
  $$\lambda = \frac{\beta_0 a}{2D} = \frac{(5 \times 10^{-3}\,\text{m})(10^{-4}\,\text{m})}{2 \times 0.5\,\text{m}} = \frac{5 \times 10^{-7}}{1.0} = 5 \times 10^{-7}\,\text{m}$$
  (or $500\,\text{nm}$).

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:
  $$\beta = \frac{(5 \times 10^{-7}\,\text{m})(2\,\text{m})}{10^{-4}\,\text{m}} = \frac{10^{-6}}{10^{-4}} = 10^{-2}\,\text{m} = 1\,\text{cm}$$

In single slit diffraction, the diffraction angle $\theta = \frac{\pi}{3} = 60^\circ$.
Using the diffraction relation $e \sin\theta = \lambda$: