The velocity of light in a transparent liquid is 1.8X10cm/second, while in vacuum it is 3X10^3cm/second. Find how much the bottom of the vessel containing this liquid appears to be raised if the depth of the liquid is 25cm.
Pretty sure light goes 3e10 cm/s in a vacuum. Is this supposed to be an optics question?
To find how much the bottom of the vessel appears to be raised, we need to consider the change in apparent depth due to the refractive index of the liquid.
The apparent depth, D', can be calculated using the formula:
D' = D/n
Where:
D is the actual depth of the liquid (given as 25 cm),
n is the refractive index of the liquid.
The refractive index (n) is the ratio of the velocity of light in vacuum (V_vacuum) to the velocity of light in the liquid (V_liquid):
n = V_vacuum / V_liquid
Given that the velocity of light in vacuum (V_vacuum) is 3 x 10^3 cm/second and the velocity of light in the liquid (V_liquid) is 1.8 x 10^8 cm/second, we can substitute these values into the refractive index equation:
n = (3 x 10^3 cm/second) / (1.8 x 10^8 cm/second)
Now we can calculate the refractive index (n):
n = 3 / 1.8 x 10^5
Next, we substitute this value of n into the formula for the apparent depth (D'):
D' = (25 cm) / (3 / 1.8 x 10^5)
Simplifying the expression:
D' = (25 cm) x (1.8 x 10^5 / 3)
D' = 25 x 1.8 x 10^5 / 3
D' = 4.5 x 10^6 cm
Therefore, the bottom of the vessel appears to be raised by 4.5 x 10^6 cm.