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Chapter 16: Problem 2244

If the critical angle for total internal reflection from a medium to vacuum is \(30^{\circ}\) then velocity of light in the medium is \(-\mathrm{ms}^{-1}\left(\right.\) take $\left.\mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) \(2.0 \times 10^{8}\) (B) \(1.5 \times 10^{8}\) (C) \(10^{8}\) (D) \(1.5 \times 10^{-8}\)

Short Answer

Expert verified
The velocity of light in the medium is \(1.5 \times 10^8 ms^{-1}\). The correct answer is (B).

Step by step solution

01

Find the refractive index of the medium

Given the critical angle for total internal reflection is 30 degrees, we can use Snell's law to find the refractive index of the medium: \[ n = \frac{\sin{\theta_2}}{\sin{\theta_1}} \] Here, \(\theta_1\) is the critical angle (30°) and \(\theta_2\) is the angle at which the refracted light ray will be at 90° (perpendicular to the interface). Plug the values: \[ n = \frac{\sin{(90^\circ)}}{\sin{(30^\circ)}} \] \[ n = \frac{1}{\frac{1}{2}} \] \[ n = 2 \]
02

Find the velocity of light in the medium

Now that we have the refractive index, we can find the velocity of light in the medium using the formula: \[ v = \frac{c}{n} \] Plug in the given speed of light in vacuum, c = \(3.0 \times 10^8 ms^{-1}\), and the refractive index, n = 2: \[ v = \frac{3.0 \times 10^8}{2} \] \[ v = 1.5 \times 10^8 ms^{-1} \] So, the velocity of light in the medium is \(1.5 \times 10^8 ms^{-1}\). The correct answer is (B).

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Most popular questions from this chapter

The light waves from two coherent sources of same intensity interfere each other. Then what will be maxima intensity when minimum intensity is zero ? (A) \(4 \mathrm{I}\) (B) I (C) \(4 \mathrm{I}^{2}\) (D) \(\mathrm{I}^{2}\)

To get five images of a single object one should have two plane mirrors at an angle of (A) \(36^{\circ}\) (B) \(72^{\circ}\) (C) \(80^{\circ}\) (D) \(302^{\circ}\)

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. (The phase difference between beam is \((\pi / 2)\) at the point \(\mathrm{A}\) and \(\pi\) at point \(\mathrm{B}\).) (A) I (B) \(4 \mathrm{I}\) (C) \(2 \mathrm{I}\) (D) \(6 \mathrm{I}\)

Ordinary light incident on a glass slab at the polarizing angle, suffers a deviation of \(22^{\circ}\). The value of angle of refraction in this case is (A) \(44^{\circ}\) (B) \(34^{\circ}\) (C) \(22^{\circ}\) (D) \(11^{\circ}\)

The fringe width for red $\beta_{\mathrm{r}}\left(\lambda_{\mathrm{T}}=8000 \AA\right.$ ) and the fringe width for violet \(\beta_{\mathrm{v}}\left(\lambda_{\mathrm{v}}=4000 \AA\right.\) ) then \(\left(\beta_{\mathrm{r}} / \beta_{\mathrm{v}}\right)=\) (A) \(2: 1\) (B) \(1: 2\) (C) \(1: 1\) (D) \(\sqrt{2}: 1\)
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