Question

A cubical glass paper-weight of side ‘a’ is placed over a physics paper. What should be the value of refractive index so that the paper can not be seen from the walls of the cube?

Answer 1

The value of refractive index so that the paper can not be seen from the walls of the cube = 1.414

Explanation:

Given: A cubical glass paper-weight of side ‘a’ is placed over a physics paper.

Find: What should be the value of refractive index so that the paper can not be seen from the walls of the cube?

Solution: We can find the required refractive index by using the concept of total internal reflection.

μ = 1/ Sinθ

μ = √2 = 1.414

Answer 2

Given that,

Side = a

According to figure,

We need to calculate the value of refractive index

Using formula of refractive index

$\dfrac{1}{\mu}=\sin\theta$

Where, $\mu$ = refractive index

Put the value into the formula

$\dfrac{1}{\mu}=\dfrac{BC}{AC}$

$\dfrac{1}{\mu}=\dfrac{a}{\sqrt{2}a}$

$\dfrac{1}{\mu}=\dfrac{1}{\sqrt{2}}$

$\mu=\sqrt{2}$

Hence, The value of refractive index is √2.

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Topic : refractive index