Question

Three immiscible transparent liquids with refractive indices \(\frac{3}{2}\), \(\frac{4}{3}\) and \(\frac{6}{5}\) are arranged one above the other. The depths of the liquids are 3 cm, 4 cm and 6 cm respectively. The apparent depth of the vessel is

• A. \(10\) cm
• B. \(9\) cm
• C. \(8\) cm
• D. \(7\) cm

Hint:

For layers viewed from above, the apparent depth is the sum of each real depth divided by its refractive index.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
When a vessel holds several immiscible layers and is viewed from above along the normal, each layer contributes a depth equal to its real depth divided by its refractive index.

Step 2: Formula
\(d_{app} = \dfrac{d_1}{\mu_1} + \dfrac{d_2}{\mu_2} + \dfrac{d_3}{\mu_3}\)

Step 3: Calculation
\(\dfrac{3}{3/2} = 2\) cm
\(\dfrac{4}{4/3} = 3\) cm
\(\dfrac{6}{6/5} = 5\) cm
Total \(= 2 + 3 + 5 = 10\) cm

Step 4: Compare the options
The apparent depth is \(10\) cm, option (A).

Final Answer:
The apparent depth of the vessel is 10 cm. \[ \boxed{\text{(A) }\text{10 cm}} \]