A glass slab of thickness 6.0 cm is placed on the piece of paper on which an inkdot is marked. By how much distance would an inkdot appear to be raised? The velocity of light in glass is 2 10^8 ms ⁻¹ and that in air is 3 10^8 ms ⁻¹ .

Step-by-step solution

The refractive index of glass is given by = c v_g = 3 10^8 2 10^8 = 1.5 = 3 2 The apparent shift of the inkdot is given by the formula x = t (1 - 1 ) Substituting the given values: x = 6.0 (1 - 2 3 ) x = 6.0 1 3 = 2.0 cm The inkdot appears to be raised by 2.0 cm . Answer: 2.0 cm