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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.

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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.

Numerical
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Solution

\[\text{Refractive index} = \frac{\text{Real depth}}{\text{Apparent depth}}\]

\[\text{Apparent depth} = \frac{\text{Real depth}}{\text{Refractive index}}\]

\[= \frac{15}{4/3}\]

= 11.25 cm

∴ Height through which image is raised = (15 − 11.25) = 3.75 cm

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Chapter 4: Refraction of Light - NUMERICAL PROBLEMS ON REFRACTION OF LIGHT THROUGH OPTICAL SLABS [Page 76]

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 10
Chapter 4 Refraction of Light
NUMERICAL PROBLEMS ON REFRACTION OF LIGHT THROUGH OPTICAL SLABS | Q 1. | Page 76
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