MHT CET · Physics · Ray Optics
A transparent glass cube of length \(24 \mathrm{~cm}\) has a small air bubble trapped inside. When seen normally through one surface from air outside, its apparent distance is \(10 \mathrm{~cm}\) from the surface. When seen normally from opposite surface, its apparent distance is \(6 \mathrm{~cm}\). The distance of the air bubble from first surface is
- A \(15 \mathrm{~cm}\)
- B \(14 \mathrm{~cm}\)
- C \(12 \mathrm{~cm}\)
- D \(8 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(15 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Given: Length of cube \(=12 \mathrm{~cm}\)
\(\mu=\frac{\text { Real depth }}{\text { Apparent depth }}=\frac{l_1}{\mathrm{~h}_1}=\frac{24-l_1}{\mathrm{~h}_2} \)
\( \text { putting } \mathrm{h}_1=10 \mathrm{~cm} \text { and } \mathrm{h}_2=6 \mathrm{~cm} \text { into (i),}\) \(\text{we get } \)
\( \frac{l_1}{10}=\frac{24-l_1}{6} \)
\( 6 l_1=240-10 l_1 \)
\( 16 l_1=240 \)
\( \therefore \quad l_1=15 \mathrm{~cm}\)
\(\mu=\frac{\text { Real depth }}{\text { Apparent depth }}=\frac{l_1}{\mathrm{~h}_1}=\frac{24-l_1}{\mathrm{~h}_2} \)
\( \text { putting } \mathrm{h}_1=10 \mathrm{~cm} \text { and } \mathrm{h}_2=6 \mathrm{~cm} \text { into (i),}\) \(\text{we get } \)
\( \frac{l_1}{10}=\frac{24-l_1}{6} \)
\( 6 l_1=240-10 l_1 \)
\( 16 l_1=240 \)
\( \therefore \quad l_1=15 \mathrm{~cm}\)
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