Question

For an anisotropic crystal of thickness 0.04 mm and refractive indices of 1.636 and 1.486 along the slow and fast directions, respectively, the retardation produced is ______ nm. (In integer)

Answer 1

Correct Answer: 6000

Calculating the Retardation Produced in an Anisotropic Crystal 

Step 1: Understanding the Formula

The retardation \( R \) produced in an anisotropic crystal is given by the formula:

\[ R = t \times (n_{\text{slow}} - n_{\text{fast}}) \]

Where:

• t is the thickness of the crystal.
• n_{\text{slow}} is the refractive index along the slow direction.
• n_{\text{fast}} is the refractive index along the fast direction.

Step 2: Given Values

Given:

• Thickness \( t = 0.04 \, \text{mm} = 0.04 \times 10^{-3} \, \text{m} \)
• Refractive index along slow direction \( n_{\text{slow}} = 1.636 \)
• Refractive index along fast direction \( n_{\text{fast}} = 1.486 \)

Step 3: Calculate the Retardation

Substitute the given values into the formula to calculate the retardation:

\[ R = 0.04 \times 10^{-3} \, \text{m} \times (1.636 - 1.486) = 0.04 \times 10^{-3} \times 0.15 = 6.0 \times 10^{-6} \, \text{m} \]

Step 4: Convert to Millimeters

Now, convert the result to millimeters:

\[ R = 6.0 \times 10^{-6} \, \text{m} \times 10^{3} = 6.0 \, \text{mm} \]

Final Answer:

The retardation produced is: 6000 (rounded to the nearest integer).