Question

A light of wavelength \(320 \mathrm{~nm}\) enters in a medium of refractive index \(1.6\) from the air of refractive index \(1.0\). The new wavelength of light in the medium will be \(\mathrm{nm}\). (A) 520 (B) 400 (C) 320 (D) 200

Short answer

The new wavelength of light in the medium is \(200 \mathrm{~nm}\) (option D).

Step-by-step solution

Identifying the formula

We will use the formula that relates the speed of light, refractive index, and wavelength of light in two different media: \(n_1 * \lambda_1 = n_2 * \lambda_2\) where, \(n_1\) and \(n_2\) are the refractive indices of the air and medium, respectively, \(\lambda_1\) and \(\lambda_2\) are the wavelengths of light in the air and medium, respectively. Our goal is to find \(\lambda_2\).

Substitute the given values

We are given the values \(n_1 = 1.0\), \(n_2 = 1.6\), and \(\lambda_1 = 320 \mathrm{~nm}\). Plug them into the formula: \(1.0 * 320 \mathrm{~nm} = 1.6 * \lambda_2\)

Solve for the new wavelength \(\lambda_2\)

Now, solve for \(\lambda_2\): \(\lambda_2 = \frac{1.0 * 320 \mathrm{~nm}}{1.6}\) \(\lambda_2 = \frac{320 \mathrm{~nm}}{1.6}\) \(\lambda_2 = 200 \mathrm{~nm}\)

Choose the correct option

From the calculation, we found the new wavelength of light in the medium to be \(200 \mathrm{~nm}\). So, the correct answer is option (D).