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Chapter 16: Problem 2228

What is the time taken in seconds to cross a glass plate of thickness $6 \mathrm{~mm}\( and \)\mu=2.0$ by light ? (A) \(8 \times 10^{-11}\) (B) \(4 \times 10^{-11}\) (C) \(2 \times 10^{11}\) (D) \(16 \times 10^{-11}\)

Short Answer

Expert verified
The time taken by light to cross the glass plate is \(4 \times 10^{-11}\, \text{s}\).

Step by step solution

01

Convert the thickness of the glass plate into meters

We are given that the thickness of the glass plate is \(6\, \text{mm}\). We need to convert this into meters. Since 1 meter is equal to 1000 millimeters, we have : \[d = 6\, \text{mm} \times \frac{1\, \text{m}}{1000\, \text{mm}}\] \[d = 0.006\, \text{m}\]
02

Use the formula to calculate the time taken by light

Now, we have the distance \(d = 0.006\, \text{m}\) and the refractive index \(\mu = 2.0\). We also know that the speed of light in vacuum is \(c = 3\times 10^8 \,\text{m/s}\). We can use the formula \(t = \frac{d}{c/\mu}\) to find the time taken by light to cross the glass plate. \[t = \frac{0.006\, \text{m}}{(3 \times 10^8 \,\text{m/s}) / 2.0}\] \[t = \frac{0.006\, \text{m}}{(1.5 \times 10^8 \,\text{m/s})}\]
03

Calculate the time and express it in scientific notation

By performing the calculation, we can find the time taken by light in seconds: \[t = \frac{0.006\, \text{m}}{(1.5 \times 10^8\, \text{m/s})}\] \[t = 4 \times 10^{-11}\, \text{s}\] The time taken by light to cross the glass plate is \(4 \times 10^{-11}\, \text{s}\). Therefore, the correct answer is (B).

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Most popular questions from this chapter

In a thin prism of glass \(\left(a_{(n) g}=1.5\right)\) which of the following relation between the angle of minimum deviation \(\delta_{\mathrm{m}}\) and the angle of refraction \(\mathrm{r}\) will be correct? (A) \(\delta_{\mathrm{m}}=(\mathrm{r} / 2)\) (B) \(\left(\delta_{\mathrm{m}} / 2\right)=\mathrm{r}\) (C) \(\delta_{\mathrm{m}}=1.5 \mathrm{r}\) (D) \(\delta_{\mathrm{m}}=\mathrm{r}\)

For four different transparent medium $\mathrm{n}_{41} \times \mathrm{n}_{12} \times \mathrm{n}_{21}=$ (A) \(\left(1 / \mathrm{n}_{41}\right)\) (B) \(\mathrm{n}_{41}\) (C) \(\mathrm{n}_{14}\) (D) \(\left(1 / \mathrm{n}_{14}\right)\)

The wave length corresponding to photon is \(0.016 \AA\). Its K.E. \(\quad\) J. $\left(\mathrm{h}=6.66 \times 10^{-34} \mathrm{SI}, \mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) \(1.237 \times 10^{-13}\) (B) \(1.237 \times 10^{13}\) (C) \(12.37 \times 10^{-13}\) (D) \(12.37 \times 10^{+13}\)

The distance between the first and sixth minima in the diffraction pattern of a single slit, it is \(0.5 \mathrm{~mm}\). The screen is \(0.5 \mathrm{~m}\) away from the Slit. If the wavelength of light is \(5000 \AA\), then the width of the slit will be \(\mathrm{mm}\) (D) \(1.0\) (A) 5 (B) \(2.5\) (C) \(1.25\)

If a ray of light is incident on a plane mirror at an angle of \(30^{\circ}\) then deviation produced by a plane mirror is (A) \(60^{\circ}\) (B) \(90^{\circ}\) (C) \(120^{\circ}\) (D) \(150^{\circ}\)
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