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 determine the critical angle for light traveling from a denser medium to air, we need to use Snell's Law and the concept of the critical angle. Snell's Law 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 media and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction. 

When light travels from one medium to air at the critical angle, \(\theta_2 = 90^\circ\) and \(\sin \theta_2 = 1\). Therefore, Snell's Law simplifies to: \(n_d \sin \theta_c = n_a\), where \(n_d\) is the refractive index of the denser medium, \(\theta_c\) is the critical angle, and \(n_a\) is the refractive index of air. Thus, \(\sin \theta_c = \frac{n_a}{n_d}\).

To find the refractive indices, we use the relation: \(n = \frac{c}{v}\), where \(c\) is the speed of light in vacuum and \(v\) is the speed of light in the medium.

Given that light travels distance \(x\) in time \(t_1\) in air, we can express the speed in air as \(v_a = \frac{x}{t_1}\). For the denser medium, the distance is \(10x\) in time \(t_2\), so the speed in the medium is \(v_d = \frac{10x}{t_2}\).

The refractive index of air is approximately \(n_a = 1\), so: \(n_d = \frac{v_a}{v_d} = \frac{\frac{x}{t_1}}{\frac{10x}{t_2}} = \frac{t_2}{10t_1}\).

Now, substituting back into the critical angle expression: \(\sin \theta_c = \frac{1}{n_d} = \frac{10t_1}{t_2}\). Therefore, the critical angle \(\theta_c\) is: \(\theta_c = \sin^{-1}\left(\frac{10t_1}{t_2}\right)\).

Answer 2

The correct option is (B): \(sin^{-1}\frac{10t_1}{t_2}\)

speed of light in air V₁=\(\frac{x}{t_1}\)

Speed of light in a medium V₂=\(\frac{10x}{t_2}\)

\(sin\theta_c=\frac{V_2}{V_1}=\frac{10x}{t_2}\frac{t_1}{x}\)

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