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A mark at the bottom of a liquid beaker ...

A mark at the bottom of a liquid beaker appears to rise by 0.1 m. If the depth of the liquid is 1.0 m, then refractive index of the liquid is ......

A

1.1

B

1.3

C

1.5

D

1.7

Text Solution

Verified by Experts

The correct Answer is:
A
`("Virtual depth h.")/("Real depth h")=(n_("air"))/(n_("water"))`
(h=1 m, h.=1-0.1=0.9 m,`n_("air")`=1)
`therefore n_w=(hxxn_("air"))/(h.)=(1xx1)/(0.9)=1.1`
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