\[ \text{Fractional error} = \frac{\text{Smallest measurement}}{\text{Measured value}} = \frac{1}{\text{Number of divisions in the circular scale}} = \frac{1}{15} \]
The error in area measurement is twice the fractional error (since the error in both dimensions contributes to the total error), so:
\[ \text{Fractional error in area} = 2 \times \frac{1}{15} = \frac{2}{15} \]
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.