Step 1: For a plano-convex lens, the approximate thickness of the lens \( t \) can be calculated using the formula: \[ t = \frac{R}{n - 1} \] where \( R \) is the radius of curvature, and \( n \) is the refractive index. Here, \( R = \frac{D}{2} = \frac{8.4}{2} = 4.2 \, \text{cm} \), and \( n = \frac{5}{3} \). Now, substituting the values: \[ t = \frac{4.2}{\frac{5}{3} - 1} = \frac{4.2}{\frac{2}{3}} = 4.2 \times \frac{3}{2} = 1.823 \, \text{cm} \] The thickness of the lens is approximately \( \boxed{1.8 \, \text{cm}} \).

A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?