A mark is made on the bottom of a vessel and over this mark, a glass slab to thickness $$3.5$$ cm and refractive index $$74$$ is placed. Now water (refractive$$ $$index.43) is poured into the vessel so that the surface of water is $$8$$ cm above the upper surface of the slab. Looking down of normally through the water, the apparent depth of the mark below the surface of water will be :

Question

A mark is made on the bottom of a vessel and over this mark, a glass slab to thickness $$3.5$$ cm and refractive index $$74$$ is placed. Now water (refractive$$ $$index.43) is poured into the vessel so that the surface of water is $$8$$ cm above the upper surface of the slab. Looking down of normally through the water, the apparent depth of the mark below the surface of water will be :

Infinity Learn · Accepted Answer

Using the relation apparent depth $$=real$$ depthμ, we get apparent depth $$=3.57/5+84/3=(2+6)cm$$ $$=$$ $$8$$ cm

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Using the relation apparent depth
$= \frac{real depth}{μ}, we get apparent depth = \frac{3.5}{7 / 5} + \frac{8}{4 / 3} = (2 + 6) cm = 8 cm$

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Ray optics

Remember concepts with our Masterclasses.

Using the relation apparent depth =real depthμ, we get apparent depth =3.57/5+84/3=(2+6)cm = 8 cm

Ready to Test Your Skills?

free study material

Using the relation apparent depth
$= \frac{real depth}{μ}, we get apparent depth = \frac{3.5}{7 / 5} + \frac{8}{4 / 3} = (2 + 6) cm = 8 cm$