Correct answer: 1.5
The refractive index \( n \) is calculated using the formula: \[ n = \frac{c}{v} \] where: - \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light in vacuum) - \( v = 2 \times 10^8 \, \text{m/s} \) (speed of light in benzene) \[ n = \frac{3 \times 10^8}{2 \times 10^8} = 1.5 \] Hence, the correct answer is 1.5.
A parallel beam of light travelling in air (refractive index \(1.0\)) is incident on a convex spherical glass surface of radius of curvature \(50 \, \text{cm}\). Refractive index of glass is \(1.5\). The rays converge to a point at a distance \(x \, \text{cm}\) from the centre of curvature of the spherical surface. The value of \(x\) is ___________.
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. 
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. 