A tiny air bubble in a glass slab (mu = 1.5) | vedclass

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(C) The refractive index $\mu$ is defined as the ratio of real depth to apparent depth: $\mu = \frac{	ext{real depth}}{	ext{apparent depth}}$.Let th

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$15$

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(C) The refractive index $\mu$ is defined as the ratio of real depth to apparent depth: $\mu = \frac{	ext{real depth}}{	ext{apparent depth}}$.Let the total thickness of the glass slab be $t$.Let the real depth of the bubble from one side be $x$. Then the real depth from the other side is $(t - x)$.Given the apparent depths are $6\, cm$ and $4\, cm$ respectively,we have:$\mu = \frac{x}{6} = 1.5 \implies x = 1.5 	imes 6 = 9\, cm$.Similarly,for the other side:$\mu = \frac{t - x}{4} = 1.5 \implies t - x = 1.5 	imes 4 = 6\, cm$.Substituting the value of $x$:$t - 9 = 6 \implies t = 15\, cm$.Therefore,the thickness of the glass slab is $15\, cm$.

$A$ tiny air bubble in a glass slab $(\mu = 1.5)$ appears from one side to be $6\, cm$ from the glass surface and from the other side,$4\, cm$. The thickness of the glass slab is......$cm$

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