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

The wavelength of light changes when it transitions between different media. The wavelength in the second medium (\( \lambda_2 \)) is related to the wavelength in the first medium (\( \lambda_1 \)) by the equation: \[ \lambda_2 = \lambda_1 \frac{v_2}{v_1} = \lambda_1 \frac{n_1}{n_2} \] Here, \( \lambda_1 \) denotes the wavelength in the initial medium (water), and \( \lambda_2 \) is the wavelength in the subsequent medium (air). Provided values: - \( \lambda_1 = 600 \) nm - \( n_1 = \frac{4}{3} \) (water) - \( n_2 = 1 \) (air) Calculation: \[ \lambda_2 = 600 \times \frac{3}{4} = 450 \text{ nm} \] Therefore, the result is: \[ \text{(B) } 450 \text{ nm} \]