A beaker contains Water and Benzene of refractive indices 43 and 32 respectively one above the other- If depth of each is 12 cm- apparent depth of the bottom of the beaker is

The correct option is B 17 cm Apparent depth= 12μ _1+12μ_2 =12 × 34+12× 23 =9+8 =17 cm ...

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Standard XII

Physics

Apparent Depth

A beaker cont...

Question

A beaker contains Water and Benzene of refractive indices $\frac{4}{3}$ and $\frac{3}{2}$ respectively one above the other. If depth of each is $12 c m$ , apparent depth of the bottom of the beaker is

17 c m

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16 c m

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19 c m

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Solution

The correct option is B $17 c m$
Apparent depth= $\frac{12}{μ_{1}} + \frac{12}{μ_{2}}$
$= \frac{12 \times 3}{4} + \frac{12 \times 2}{3}$
$= 9 + 8$
$= 17 c m$

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A beaker contains Water and Benzene of refractive indices 43 and 32 respectively one above the other. If depth of each is 12 cm, apparent depth of the bottom of the beaker is

The correct option is B 17 cmApparent depth= 12μ1+12μ2=12×34+12×23=9+8=17cm

A beaker contains Water and Benzene of refractive indices $\frac{4}{3}$ and $\frac{3}{2}$ respectively one above the other. If depth of each is $12 c m$ , apparent depth of the bottom of the beaker is

The correct option is B $17 c m$
Apparent depth= $\frac{12}{μ_{1}} + \frac{12}{μ_{2}}$
$= \frac{12 \times 3}{4} + \frac{12 \times 2}{3}$
$= 9 + 8$
$= 17 c m$