A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index $$\dfrac{5}{3}$$ is poured inside the bucket, then microscope have to be raised by 30 cm to focus the object again. The height of the liquid in the bucket is:

A microscope focused on an object at the bottom of a bucket must be raised 30 cm when a liquid of refractive index 5/3 is poured in. We need the height of the liquid.

Understand apparent depth and real depth.

When light travels from a denser medium (liquid) to a rarer medium (air), the apparent position of an object is closer to the surface than its actual position. The relationship is:

$$ \text{Apparent depth} = \frac{\text{Real depth}}{n} $$

where $$n$$ is the refractive index of the liquid.

Calculate the apparent shift.

The apparent shift (the amount the object appears to rise) is:

$$ \text{Shift} = \text{Real depth} - \text{Apparent depth} = h - \frac{h}{n} = h\left(1 - \frac{1}{n}\right) $$

This is the distance the microscope must be raised to refocus.

Substitute and solve.

Given: shift = 30 cm, $$n = 5/3$$:

$$ 30 = h\left(1 - \frac{1}{5/3}\right) = h\left(1 - \frac{3}{5}\right) = h \times \frac{2}{5} $$

$$ h = \frac{30 \times 5}{2} = 75\;\text{cm} $$

The height of the liquid is 75 cm.

The correct answer is Option 1: 75 cm.