KCET · Physics · Ray Optics
The time required for the light to pass through a glass slab (refractive index \(=1.5\) ) of thickness \(4 \mathrm{~mm}\) is
( \(c=3 \times 10^{8} \mathrm{~ms}^{-1}\), speed of light in free space)
- A \(10^{-11} \mathrm{~s}\)
- B \(2 \times 10^{-11} \mathrm{~s}\)
- C \(2 \times 10^{11} \mathrm{~s}\)
- D \(2 \times 10^{-5} \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(2 \times 10^{-11} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
We know,
\(
\begin{gathered}
n_{a} c_{a}=n_{g} c_{g} \\
\frac{n_{g}}{n_{a}}=\frac{c_{a}}{c_{g}} \\
\frac{3}{2}=\frac{3 \times 10^{8}}{c_{g}} \\
c_{g}=2 \times 10^{8}
\end{gathered}
\)
We have, \(\quad\) Time \(=\frac{\text { Distance }}{\text { Speed }}\)
\(t=\frac{4 \times 10^{-3}}{2 \times 10^{8}}\)
or
\(t=2 \times 10^{-11} \mathrm{~s}\)
\(
\begin{gathered}
n_{a} c_{a}=n_{g} c_{g} \\
\frac{n_{g}}{n_{a}}=\frac{c_{a}}{c_{g}} \\
\frac{3}{2}=\frac{3 \times 10^{8}}{c_{g}} \\
c_{g}=2 \times 10^{8}
\end{gathered}
\)
We have, \(\quad\) Time \(=\frac{\text { Distance }}{\text { Speed }}\)
\(t=\frac{4 \times 10^{-3}}{2 \times 10^{8}}\)
or
\(t=2 \times 10^{-11} \mathrm{~s}\)
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