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Question

A ring is placed at the bottom of a trough containing four immiscible liquids of refractive indices $1.0, 2.0, 3.0$ and $4.0$ poured one above the other of heights $10$ cm, $20$ cm, $30$ cm, and $40$ cm respectively. The apparent depth at which the ring appears, when seen from outside is____________ cm.

Apparent Depth of Ring Calculation

This problem involves calculating the apparent depth of an object (a ring) at the bottom of a container filled with multiple immiscible liquids. The apparent depth differs from the real depth due to the refraction of light as it travels from the liquid to the observer's eye in the air.

Apparent Depth Physics Formula

When observing from air (refractive index $n_{air} \approx 1$) into a single liquid layer of depth $h$ and refractive index $n$, the apparent depth $d_{app}$ is calculated as: $ d_{app} = \frac{h}{n} $ For multiple liquid layers stacked one above the other, the total apparent depth ($D_{app}$) is the sum of the apparent depths of each individual layer. The formula becomes: $ D_{app} = \sum_{i=1}^{N} \frac{h_i}{n_i} $ where $h_i$ is the real depth of the $i$-th layer and $n_i$ is its refractive index.

Trough Liquid Layers Application

The given information is:

Layer Real Depth ($h_i$) Refractive Index ($n_i$)
1 10 cm 1.0
2 20 cm 2.0
3 30 cm 3.0
4 40 cm 4.0

Calculating Ring's Apparent Depth

Apply the formula for total apparent depth using the values provided:

$ D_{app} = \frac{h_1}{n_1} + \frac{h_2}{n_2} + \frac{h_3}{n_3} + \frac{h_4}{n_4} $

Substitute the depths and refractive indices:

$ D_{app} = \frac{10 \text{ cm}}{1.0} + \frac{20 \text{ cm}}{2.0} + \frac{30 \text{ cm}}{3.0} + \frac{40 \text{ cm}}{4.0} $

Perform the division for each term:

$ D_{app} = 10 \text{ cm} + 10 \text{ cm} + 10 \text{ cm} + 10 \text{ cm} $

Sum the results to find the total apparent depth:

$ D_{app} = 40 \text{ cm} $

The ring appears at an apparent depth of 40 cm from the surface of the top liquid layer.

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  1. In a microscope the objective is having focal length $f_o = 2 \text{ cm}$ and eye-piece is having focal length $f_e = 4 \text{ cm}$. The tube length is $32 \text{ cm}$. The magnification produced by this microscope for normal adjustment is ____________.
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