Question

A ray of light of wavelength 600 nm is incident in water (\( n = \frac{4}{3} \)) on the water-air interface at an angle less than the critical angle. The wavelength associated with the refracted ray is:

Answer 1

When light passes from one medium to another, the wavelength of the light changes. The relationship between the wavelengths in the two media is given by: \[ \lambda_2 = \lambda_1 \frac{v_2}{v_1} = \lambda_1 \frac{n_1}{n_2} \] where \( \lambda_1 \) is the wavelength in the first medium (water), and \( \lambda_2 \) is the wavelength in the second medium (air). Given: - \( \lambda_1 = 600 \) nm - \( n_1 = \frac{4}{3} \) (water) - \( n_2 = 1 \) (air) Thus: \[ \lambda_2 = 600 \times \frac{3}{4} = 450 \text{ nm} \] Thus, the correct answer is: \[ \text{(B) } 450 \text{ nm} \]