Sum

Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.

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#### Solution

Given,

Taking the radius of the paperweight as its thickness = 3 cm

Refractive index of the paperweight (μ_{g}) = 3/2

Refractive index of the air (μ_{1}) = 1

Image shift is given by:

\[∆ t = \left( 1 - \frac{1}{\mu} \right)t\]

\[ = \left( 1 - \frac{2}{3} \right)3\]

\[ = \left( \frac{1}{3} \right)3\]

\[ = 1 \text{ cm }\]

The upper surface of the paperweight is flat and the spherical spherical surface is in contact with the printed letter.

Therefore, we will take it as a simple refraction problem.

Hence, the image will appear 1 cm above point A.

Concept: Total Internal Reflection

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