Refractive index of a medium nm is given by,
\(n_m = \frac{\text{Speed of light in vacuum}}{\text{Speed of light in the medium }}\)
\(=\frac{ c }{v}\)
Speed of light in vacuum, c = 3 × 108 m s−1
Refractive index of glass, ng = 1.50
Speed of light in the glass,
\(v=\frac{c}{n_g}\)
\(=\frac{3\times10^8}{1.50}\)
\(=2\times10^8\) ms-1
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, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), 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.