A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised, if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? (Refraction index of glass = 1.5). Does the answer depend on the location of the slab?

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Solution

We know that $η = \frac{r e a l d i s t a n c e}{a p p a r e n t d i s t a n c e}$
$1.5 = \frac{15}{x}$
$x = 10 c m$
So it is shifted up by $15 - 10 = 5 c m$ .
No the answer does not depend on location of slap.

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We know that η=realdistanceapparentdistance1.5=15xx=10cm So it is shifted up by 15−10=5cm.No the answer does not depend on location of slap.

We know that $η = \frac{r e a l d i s t a n c e}{a p p a r e n t d i s t a n c e}$
$1.5 = \frac{15}{x}$
$x = 10 c m$
So it is shifted up by $15 - 10 = 5 c m$ .
No the answer does not depend on location of slap.