Question

A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height $ H $ above the first liquid. If viewed from above, the apparent shift in the position of the bottom of the container is 40 cm. The value of $ H $ is ___ cm.

Hint:

When dealing with problems involving apparent depth and refraction, use the formula \( \text{Apparent depth} = \frac{\text{Real depth}}{\text{Refractive index}} \) and apply it to the layers of liquids with different refractive indices.

Answer 1

Correct Answer: 80

Step 1: Understand Apparent Shift
The apparent shift occurs due to refraction at the liquid-air interface. The total apparent shift is the sum of shifts caused by each liquid layer.
Step 2: Formula for Apparent Shift
For a liquid layer of height \( h \) and refractive index \( \mu \), the apparent shift is given by: \[ \text{Shift} = h\left(1 - \frac{1}{\mu}\right) \]
Step 3: Calculate Shifts for Both Liquids

• For the first liquid (\( \mu_1 = 1.2 \), \( h_1 = 60 \) cm): \[ \text{Shift}_1 = 60\left(1 - \frac{1}{1.2}\right) = 60\left(1 - \frac{5}{6}\right) = 60 \times \frac{1}{6} = 10 \text{ cm} \]
• For the second liquid (\( \mu_2 = 1.6 \), \( h_2 = H \) cm): \[ \text{Shift}_2 = H\left(1 - \frac{1}{1.6}\right) = H\left(1 - \frac{5}{8}\right) = H \times \frac{3}{8} = \frac{3H}{8} \text{ cm} \]

Step 4: Total Apparent Shift
The total apparent shift is given as 40 cm: \[ \text{Shift}_1 + \text{Shift}_2 = 40 \text{ cm} \] \[ 10 + \frac{3H}{8} = 40 \]
Step 5: Solve for \( H \)
\[ \frac{3H}{8} = 30 \] \[ 3H = 240 \] \[ H = 80 \text{ cm} \]
Verification
Let's verify the calculation: \[ \text{Total shift} = 10 + \frac{3 \times 80}{8} = 10 + 30 = 40 \text{ cm} \] This matches the given condition.
Conclusion
The height \( H \) of the second liquid layer is \(80\) cm.