Question

With respect to air, the critical angle in a medium for red light of wave length $ \lambda_1 $ is $ \theta $. Other facts remaining same, the critical angle for yellow light of wave length $ \lambda_2 $ will be Options:

• A. \( \theta \)
• B. more than \( \theta \)
• C. less than \( \theta \)
• D. \( \frac{\theta \lambda_1}{\lambda_2} \)

Hint:

The critical angle is dependent on the refractive index, which varies with the wavelength of light. Shorter wavelengths (like yellow light) lead to a smaller critical angle.

Answer 1

The Correct Option is C

1. The critical angle for a wave is the angle of incidence beyond which total internal reflection occurs. The formula for the critical angle \( \theta_c \) is given by: \[ \sin \theta_c = \frac{n_2}{n_1} \] where \( n_1 \) is the refractive index of the medium from which the light is coming (air, in this case), and \( n_2 \) is the refractive index of the medium in which the light is refracted.
2. The refractive index \( n_2 \) of a medium for different wavelengths (colors of light) varies. For shorter wavelengths (such as violet light), the refractive index is higher, and for longer wavelengths (such as red light), the refractive index is lower.
3. Given that \( \lambda_2 \) (yellow light) is shorter than \( \lambda_1 \) (red light), it has a higher refractive index \( n_2 \) for the same medium.
4. Since \( \sin \theta_c \) is inversely proportional to the refractive index, a higher refractive index results in a smaller critical angle.
5. Therefore, the critical angle for yellow light \( \lambda_2 \) will be less than the critical angle for red light \( \lambda_1 \). \[ \boxed{\text{less than } \theta} \]