Question:
A glass slab of thickness \(6.0\) cm is placed on the piece of paper on which an inkdot is marked. By how much distance would an inkdot appear to be raised? The velocity of light in glass is \(2\times 10^8\text{ ms}^{-1}\) and that in air is \(3\times 10^8\text{ ms}^{-1}\).
A glass slab of thickness \(6.0\) cm is placed on the piece of paper on which an inkdot is marked. By how much distance would an inkdot appear to be raised? The velocity of light in glass is \(2\times 10^8\text{ ms}^{-1}\) and that in air is \(3\times 10^8\text{ ms}^{-1}\).
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Apparent shift is t(1 - 1/mu), with mu = c/v.
Updated On: Oct 1, 2026
- \(2.0\) cm
- \(3.0\) cm
- \(4.0\) cm
- \(5.0\) cm
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The Correct Option is A
Solution and Explanation
Step 1: Refractive Index:
\[ \mu=\frac{c}{v}=\frac{3\times10^8}{2\times10^8}=1.5 \]
Step 2: Normal Shift:
For a slab of thickness \(t\), viewed from above, the apparent shift is
\[ \text{shift}=t\left(1-\frac1\mu\right)=6\left(1-\frac1{1.5}\right)=6\times\frac13=2.0\ \text{cm} \]
Step 3: Check the Other Options:
3.0 cm would need \(\mu=2\), 4.0 cm would need \(\mu=3\), and 5.0 cm would need \(\mu=6\). With \(\mu=1.5\), only 2.0 cm fits.
Final Answer:
The ink dot appears raised by 2.0 cm, option (A). \[ \boxed{\text{(A) } 2.0\ \text{cm}} \]
\[ \mu=\frac{c}{v}=\frac{3\times10^8}{2\times10^8}=1.5 \]
Step 2: Normal Shift:
For a slab of thickness \(t\), viewed from above, the apparent shift is
\[ \text{shift}=t\left(1-\frac1\mu\right)=6\left(1-\frac1{1.5}\right)=6\times\frac13=2.0\ \text{cm} \]
Step 3: Check the Other Options:
3.0 cm would need \(\mu=2\), 4.0 cm would need \(\mu=3\), and 5.0 cm would need \(\mu=6\). With \(\mu=1.5\), only 2.0 cm fits.
Final Answer:
The ink dot appears raised by 2.0 cm, option (A). \[ \boxed{\text{(A) } 2.0\ \text{cm}} \]
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