Question

a coin is placed at a depth of 15 CM in a beaker containing water .The refractive index of water is 4/3 .Calculate height through which the image of the coin is raised.
The answer is (3.75)​

Answer 1

Answer:

as u know the formula

hope u understand

Answer 2

Concept:

Refractive index of glass is defined as the ratio of apparent depth to the real depth which is given by,

$\frac{u_{2} }{u_{1} }=\frac{apparent depth}{real depth}$

Given:

Refractive index of water, $\frac{u_{2} }{u_{1} }=\frac{3}{4}$

Real depth, $15cm$

To Find:

We have to find the height through which the image of the coin is raised.

Solution:

Firstly, we find the apparent depth

by considering the given data,

$\frac{u_{2} }{u_{1} }=\frac{apparent depth}{real depth}=\frac{x}{15}$

$\frac{3}{4}=\frac{x}{15}$

$x=11.25$

height $\delta x$ through which the image of the coin is raised

$\delta x=15-11.25=3.75$

Hence the height through which the image of the coin is raised is $3.75$

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