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Chapter 16: Problem 2212

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

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

Step by step solution

01

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 \(\lambda_2\).
02

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: \(1.0 * 320 \mathrm{~nm} = 1.6 * \lambda_2\)
03

Solve for the new wavelength \(\lambda_2\)

Now, solve for \(\lambda_2\): \(\lambda_2 = \frac{1.0 * 320 \mathrm{~nm}}{1.6}\) \(\lambda_2 = \frac{320 \mathrm{~nm}}{1.6}\) \(\lambda_2 = 200 \mathrm{~nm}\)
04

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).

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