A glass tumbler having an inner depth of 17.5 | vedclass

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(B) Let the actual height of the water in the tumbler be $H$. The refractive index of water is $\mu_w = 4/3$.The apparent depth of the water as seen b

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$10$

Vedclass · Answer

(B) Let the actual height of the water in the tumbler be $H$. The refractive index of water is $\mu_w = 4/3$.The apparent depth of the water as seen by the observer from above is given by $d_{app} = \frac{H}{\mu_w} = \frac{H}{4/3} = \frac{3H}{4}$.The height of the empty part of the tumbler (air column) is $17.5 - H$.According to the problem,the student feels the tumbler is half-filled,which means the apparent depth of the water appears equal to the height of the empty part of the tumbler.Therefore,$\frac{3H}{4} = 17.5 - H$.Adding $H$ to both sides: $\frac{3H}{4} + H = 17.5$.$\frac{7H}{4} = 17.5$.$H = \frac{17.5 	imes 4}{7} = 2.5 	imes 4 = 10 \, cm$.

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