A glass slab of thickness $12$ mm is placed on a table. The lower surface of the slab has a black spot. At what depth in (mm) from upper surface will the spot appear, when viewed from above $(μ = 1.5)$ .

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Solution

Refractive index $= \frac{r e a l d e p t h}{a p p a r e n t d e p t h}$
$1.5 = \frac{12}{a p p a r e n t d e p t h}$
$a p p a r e n t d e p t h = \frac{12}{1.5} = 8 m m$

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A glass slab of thickness 12 mm is placed on a table. The lower surface of the slab has a black spot. At what depth in (mm) from upper surface will the spot appear, when viewed from above (μ=1.5).

Refractive index=realdepthapparentdepth1.5=12apparentdepth apparentdepth=121.5=8mm

A glass slab of thickness $12$ mm is placed on a table. The lower surface of the slab has a black spot. At what depth in (mm) from upper surface will the spot appear, when viewed from above $(μ = 1.5)$ .

Refractive index $= \frac{r e a l d e p t h}{a p p a r e n t d e p t h}$
$1.5 = \frac{12}{a p p a r e n t d e p t h}$
$a p p a r e n t d e p t h = \frac{12}{1.5} = 8 m m$