Question

A travelling microscope is focused on to a scratch on the bottom of a beaker. Water of refractive index 4/3 is poured in it. Then the microscope is to be lifted through 2 cm focus it again. Find the depth of water in the beaker.
(1) 4 cm
(2) 8 cm
(3) 8/7 cm
(4) 8/3 cm

Answer 1

Answer on Question #38974, Physics, Optics

A travelling microscope is focused on to a scratch on the bottom of a beaker. Water of refractive index $4/3$ is poured in it. Then the microscope is to be lifted through $2\mathrm{cm}$ focus it again. Find the depth of water in the beaker.

(1) $4 \mathrm{~cm}$

(2) (2) $8 \mathrm{~cm}$

(3) (3) $8 / 7 \mathrm{~cm}$

(4) (4) $8 / 3 \mathrm{~cm}$

Solution:

Apparent depth is depth that a transparent material such as water appears to have when viewed from above. This is less than its real depth because of the refraction that takes place when light passes into a less dense medium.

$$d _ {r e a l} - d _ {a p p} = 2 \mathrm {c m}$$

The ratio of the real depth to the apparent depth of a transparent material is equal to its refractive index.

Refractive index $n =$ Real depth/Apparent depth.

Thus,

$$n = \frac {d _ {r e a l}}{d _ {a p p}} = \frac {4}{3}$$

We have system of equation

$$\left\{ \begin{array}{c} d _ {r e a l} - d _ {a p p} = 2 \\ \frac {d _ {r e a l}}{d _ {a p p}} = \frac {4}{3} \end{array} \right.$$

A solution of a system is

$$d _ {a p p} = \frac {3}{4} d _ {r e a l}$$ $$d _ {r e a l} - \frac {3}{4} d _ {r e a l} = 2$$ $$\frac {1}{4} d _ {r e a l} = 2$$ $$d _ {r e a l} = 8$$

Answer. (2) $8 \mathrm{~cm}$ .