Consider the situation as shown in figure. A plane mirror is fixed at a height h above the bottom of a beaker containing water (refractive index μ ) up to a height d. Find the position of the image of the bottom formed by the mirror.
A
h−d+dμ
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B
h+d+dμ
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C
h−d−dμ
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D
d+dμ
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Solution
The correct option is Ah−d+dμ The bottom of the beaker appears to be shifted up by a distance △t=(1−1μ)d
Thus, the apparent distance of the bottom from the mirror is h−△t=h−(1−1μ)d=h−d+dμ. The image is formed behind the mirror at a distance h−d+dμ.