Question

For a crane, apparent depth of fish is 6cm. what will be the real depth
Given: Refractive index of water(n)=4/3​

Answer 1

$d_r=8\ cm$ is the real depth of the object

Explanation:

Using Snell's law:

Refractive index:

$n=\frac{d_r}{d_a}$ ............................(1)

where

$d_r\ \&\ d_a$ are real and apparent depths respectively.

Using eq. (1)

$\frac{4}{3} =\frac{d_r}{6}$

$d_r=8\ cm$

here we are seeing from air into the water which is optically denser medium.

TOPIC: optical density, refractive index

#LearnMore

https://brainly.in/question/10517843

Answer 2

The real depth of the fish is 8 cm.

Explanation:

It is given that,

Apparent depth of the fish, d' = 6 cm

Refractive index of water, n = 4/3

We need to find the real depth of the fish. The refractive index of any medium is given in terms of real and apparent depth is given by :

$n=\dfrac{real\ depth}{apparent\ depth}$

$n=\dfrac{d}{d'}$

$d=n\times d'$

$d=\dfrac{4}{3}\times 6$

d = 8 cm

So, the real depth of the fish is 8 cm. Hence, this is the required solution.

Learn more,

For a crane, apparent depth of fish is 6 cm. What will be the real depth?

https://brainly.in/question/10517843