Question

A beaker is filled with water (refractive index \( \frac{4}{3} \)) up to a height \( H \). A coin is placed at its bottom. The depth of the coin, when viewed along the near normal direction, will be:

• A. \( 3H \)
• B. \( \frac{3H}{4} \)
• C. \( H \)
• D. \( \frac{4H}{3} \)

Hint:

The apparent depth of an object in a transparent medium is given by dividing the real depth by the refractive index of the medium.

Answer 1

The Correct Option is B

To determine the apparent depth of a coin placed at the bottom of a beaker filled with water, we use the concept of the refractive index and Snell's law. The apparent depth (\(d_a\)) is given by the formula: 

\(d_a = \frac{d}{n}\)

where \(d\) is the actual depth of the coin and \(n\) is the refractive index of the medium (water in this case).

Given:

• Actual depth, \(d = H\)
• Refractive index of water, \(n = \frac{4}{3}\)

Substituting the values, we find:

\(d_a = \frac{H}{\frac{4}{3}}\)

To solve this, multiply \(H\) by the reciprocal of \(\frac{4}{3}\):

\(d_a = H \times \frac{3}{4} = \frac{3H}{4}\)

Therefore, the depth of the coin, when viewed along the near normal direction, is \( \frac{3H}{4} \).