Question

Linearly polarized light (free space wavelength 𝜆₀= 600nm) is incident normally on a retarding plate (𝑛_𝑒-𝑛_𝑜 = 0.05 at 𝜆₀= 600nm). The emergent light is observed to be linearly polarized, irrespective of the angle between the direction of polarization and the optic axis of the plate. The minimum thickness (in µm) of the plate is:

• A. 6
• B. 3
• C. 2
• D. 1

Answer 1

The Correct Option is A

The phase difference \( \Delta \phi \) between the ordinary and extraordinary rays is given by: \[ \Delta \phi = \frac{2\pi}{\lambda_0} (n_e - n_o) \, d. \] For the light to remain linearly polarized, the phase difference must be a multiple of \( 2\pi \): \[ \Delta \phi = 2\pi m \quad \text{(where \( m \) is an integer)}. \]

Substituting \( \Delta \phi = 2\pi \) for the minimum thickness (i.e., \( m = 1 \)): \[ \frac{2\pi}{\lambda_0} (n_e - n_o) \, d = 2\pi. \] Simplify: \[ \frac{(n_e - n_o) \, d}{\lambda_0} = 1. \] Rearrange for \( d \): \[ d = \frac{\lambda_0}{n_e - n_o}. \]

Step 2: Substitute the values 

Given: \[ \lambda_0 = 600 \, \text{nm} = 0.6 \, \mu\text{m}, \quad n_e - n_o = 0.05. \] Substitute into the formula: \[ d = \frac{0.6}{0.05} = 12 \, \mu\text{m}. \]

Step 3: Minimum Thickness

The plate’s thickness must account for half-wavelength retardation for the given wavelength, dividing \( d \) by 2: \[ d_{\text{min}} = \frac{12}{2} = 6 \, \mu\text{m}. \]

Final Answer:

The minimum thickness of the plate is 6 µm.