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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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Most popular questions from this chapter

Read the paragraph and chose the correct answer of the following questions In young experiment position of bright fringes is given by $\mathrm{x}=\mathrm{n} \lambda(\mathrm{D} / \mathrm{d})$ and the position of dark fringes is given by \(\mathrm{x}=(2 \mathrm{n}-1)(N 2)(\mathrm{D} / \mathrm{d})\) where \(\mathrm{n}=1,2,3 \ldots \ldots \ldots \ldots\) for first second, third bright/dark fringe. The center of the fringe pattern is bright (for \(\mathrm{n}=0\) ). The width of each bright/dark fringe is $\beta=(\lambda \mathrm{D} / \mathrm{d})\(, Where \)\lambda=5000 \AA\(. Slits are \)0.2 \mathrm{~cm}$ apart and \(\mathrm{D}=1 \mathrm{~m}\) (i) If light of wavelength \(6000 \AA\) be used in the above experiment the fringe width would be \(\mathrm{mm}\) (A) \(0.30\) (B) 3 (C) \(0.6\) (D) 6 (ii) with the light of wavelength \(5000 \AA\), If experiment were carried out under water of a \(n=(4 / 3)\) the fringe width would be (A) zero (B) \((4 / 3)\) times (C) (3/4) times (D) none of these

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