Question

Light travels a distance x in time t1 in air and 10x in time t2 in another denser medium. What is the critical angle for this medium?

• A. sin-¹(\(\frac{10t_1}{t_2}\))
• B. sin^-1(\(\frac{t_2}{t_1}\))
• C. sin^-1(1\(\frac{10t_2}{t_1}\))
• D. sin^-1(\(\frac{t_1}{10t_2}\))

Answer 1

The Correct Option is A

To find the critical angle for light traveling from a denser medium to air, we can use Snell's Law, which is given by:

\(n_1 \sin \theta_1 = n_2 \sin \theta_2\)

Where:

• \(n_1\) and \(n_2\) are the refractive indices of the first and second medium respectively.
• \(\theta_1\) is the angle of incidence.
• \(\theta_2\) is the angle of refraction.

The critical angle, \(\theta_c\), occurs when the angle of refraction is \(90^\circ\), meaning:

\(\sin \theta_c = \frac{n_2}{n_1}\)

In this problem, we are given the distances traveled by light in air and the denser medium:

• Distance in air: \(x\), Time: \(t_1\)
• Distance in denser medium: \(10x\), Time: \(t_2\)

We can find the speed of light in each medium:

• Speed in air, \(v_{\text{air}} = \frac{x}{t_1}\)
• Speed in denser medium, \(v_{\text{medium}} = \frac{10x}{t_2}\)

Refractive index \(n\) is inversely proportional to speed, so the refractive index of air \((n_{\text{air}} = 1)\):

• \(n_{\text{medium}} = \frac{v_{\text{air}}}{v_{\text{medium}}} = \frac{x/t_1}{10x/t_2} = \frac{t_2}{10t_1}\)

Now calculate the critical angle \(\theta_c\) using:

\(\sin \theta_c = \frac{n_{\text{air}}}{n_{\text{medium}}} = \frac{1}{n_{\text{medium}}} = \frac{10t_1}{t_2}\)

Thus, the critical angle is:

\(\theta_c = \sin^{-1}\left(\frac{10t_1}{t_2}\right)\)

Therefore, the correct answer is the option: \(\sin^{-1}\left(\frac{10t_1}{t_2}\right)\).