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Question: An air bubble in a glass sphere having 4 *cm* diameter appears 1 *cm* from surface nearest to eye wh...

An air bubble in a glass sphere having 4 cm diameter appears 1 cm from surface nearest to eye when looked along diameter. If aμg=1.5{ } _ { a } \mu _ { g } = 1.5 , the distance of bubble from refracting surface is

A

1.2 cm

B

3.2 cm

C

2.8 cm

D

1.6 cm

Answer

1.2 cm

Explanation

Solution

By using

μ2v−μ1u=μ2−μ1R\frac { \mu _ { 2 } } { v } - \frac { \mu _ { 1 } } { u } = \frac { \mu _ { 2 } - \mu _ { 1 } } { R }

where u = ? , v = – 1 cm, μ1=1.5\mu _ { 1 } = 1.5, μ2=1\mu _ { 2 } = 1 ,

R = – 2 cm.

1−1−1.5u=1−1.5(−2)\frac { 1 } { - 1 } - \frac { 1.5 } { u } = \frac { 1 - 1.5 } { ( - 2 ) }

⇒ u=−65=−1.2 cmu = - \frac { 6 } { 5 } = - 1.2 \mathrm {~cm}