A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
The pin appears to be raised by 5 cm. The answer does not depend on the location of the slab.
Explanation
This question applies the concept of apparent depth due to refraction. The textbook states that 'the apparent depth (h₁) is real depth (h₂) divided by the refractive index of the medium.' For a glass slab, the shift depends only on thickness and refractive index, not position.
Solution Steps
Step 1: Identify given values - Real depth (thickness of glass slab) = 15 cm, Refractive index of glass (n) = 1.5
Step 2: Apply the formula for apparent depth: Apparent depth = Real depth / n = 15 / 1.5 = 10 cm
Step 3: Calculate the shift (distance by which pin appears raised): Shift = Real depth - Apparent depth = 15 - 10 = 5 cm
Step 4: For the second part - The apparent shift depends only on the thickness of the slab and the refractive index. The formula does not involve the position of the slab. Hence, the answer does not depend on the location of the slab.