Physics 2002 JAMB Past Questions - Question 14: A coin placed below a rectangular glass block of thickness 9 cm and refractive index 1.5 is viewed vertically above tie block. The apparent displacement of the coin is

Correct Answer

C

Explanation

n=real depth/apparent depth; Apparent depth=6cmApparent displacement =9-6=3cmWhen a coin is placed below a rectangular glass block with a refractive index greater than 1 (in this case, n = 1.5), and you view it vertically above the block, it appears to be displaced laterally due to refraction. This lateral displacement is often called "apparent displacement."

To calculate the apparent displacement, you can use the following formula:

Apparent Displacement (D) = t * (n - 1)

Where:
- D is the apparent displacement of the object (coin).
- t is the thickness of the glass block.
- n is the refractive index of the glass.

In this case:
- t (thickness of the glass block) = 9 cm = 0.09 meters (since 1 cm = 0.01 meters).
- n (refractive index of the glass) = 1.5

Now, plug these values into the formula:

D = 0.09 meters * (1.5 - 1)
D = 0.09 meters * 0.5
D = 0.045 meters

So, the apparent displacement of the coin when viewed vertically above the glass block is 0.045 meters, or 4.5 centimeters.