A coin is placed at the bottom of a beaker containing water (refractive index $= 4 / 3$ ) to a depth of $12 c m$ . By what height the coin appears to be raised when seen from vertically above?

9 c m

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16 c m

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3 c m

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4 c m

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Solution

The correct option is C $3 c m$
$R I = \frac{real depth}{apparent depth}$

Substituting, $\frac{4}{3} = \frac{12}{a p p a r e n t d e p t h}$

Therefore, $a p p a r e n t d e p t h = \frac{12 \times 3}{4} = 9$

The height by which it appears to be raised is $12 - 9 = 3 c m$

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A coin is placed at the bottom of a beaker containing water (refractive index =4/3) to a depth of 12cm. By what height the coin appears to be raised when seen from vertically above?

The correct option is C 3 cmRI=real depthapparent depthSubstituting, 43=12apparentdepthTherefore, apparent depth=12×34=9The height by which it appears to be raised is 12−9=3cm

A coin is placed at the bottom of a beaker containing water (refractive index $= 4 / 3$ ) to a depth of $12 c m$ . By what height the coin appears to be raised when seen from vertically above?

The correct option is C $3 c m$
$R I = \frac{real depth}{apparent depth}$

Substituting, $\frac{4}{3} = \frac{12}{a p p a r e n t d e p t h}$

Therefore, $a p p a r e n t d e p t h = \frac{12 \times 3}{4} = 9$

The height by which it appears to be raised is $12 - 9 = 3 c m$