Question

An object at the bottom of a beaker filled with a liquid up to height of 10cm. If therefractive index of liquid with respect to air is 4/3, find the apparent depth of the object.​

Answer 1

Answer:

Given:

Height (h) = 10 cm

Refractive Index (μ) = 4/3

To Find:

The apparent depth (d) of the object

Solution:

We know that,

Apparent depth is the depth which is visible to human eye due to refraction of light in different medium.

The formula for Apparent depth (d)

$\sf Apparent\;depth = \dfrac{Real\;depth}{Refractive\;index}$

Mathematically,

$\blue{\boxed{\boxed{\sf d = \dfrac{h}{\mu} }}}$

Where,

h = Real Height = 10 cm

μ = Refractive Index = 4/3

Therefore,

Substituting the above values in the given formula

We get,

$\sf \implies d = \dfrac{10}{\dfrac{4}{3} }\;cm$

$\sf \implies d = 10 \times \dfrac{3}{4}\;cm$

$\sf \implies d = \dfrac{30}{4}\;cm$

$\sf \implies d = 7.5\;cm$

$\green{\boxed{\boxed{\sf d = 7.5\;cm}}}$

Hence

The apparent depth of the object is 7.5 cm

Answer 2

Explanation:

Answer:

Given:

Height (h) = 10 cm

Refractive Index (μ) = 4/3

To Find:

The apparent depth (d) of the object

Solution:

We know that,

Apparent depth is the depth which is visible to human eye due to refraction of light in different medium.

The formula for Apparent depth (d)

\sf Apparent\;depth = \dfrac{Real\;depth}{Refractive\;index}Apparentdepth=

Refractiveindex

Realdepth

Mathematically,

\blue{\boxed{\boxed{\sf d = \dfrac{h}{\mu} }}}

d=

μ

h

Where,

h = Real Height = 10 cm

μ = Refractive Index = 4/3

Therefore,

Substituting the above values in the given formula

We get,

\sf \implies d = \dfrac{10}{\dfrac{4}{3} }\;cm⟹d=

3

4

10

cm

\sf \implies d = 10 \times \dfrac{3}{4}\;cm⟹d=10×

4

3

cm

\sf \implies d = \dfrac{30}{4}\;cm⟹d=

4

30

cm

\sf \implies d = 7.5\;cm⟹d=7.5cm

\green{\boxed{\boxed{\sf d = 7.5\;cm}}}

d=7.5cm