Thickness of Lead to block photons

$\frac{h _1}{84.56 cm} = tan(1.95 \times .25 \cdot 1.95) =\gt h _1 = 2.1589 cm$

$\frac{h _2}{84.56 + d} = tan(1.95 + .25 \cdot 1.95) =\gt h _2 = .042568d + 3.599557$

$tan \theta = \frac{h _1}{4.56 cm} =\gt 25.6^{\circ}$

$tan \theta = \frac{h _2}{4.56 cm + d} =\gt tan 25.6 =\frac{.042568d + 3.599557}{d + 4.56 cm}$

∴ $\textbf{the}$ $\textbf{thickness}$ $\textbf{of}$ $\textbf{the}$ $\textbf{lead = 3.24 cm}$