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Chapter 16: Problem 2242

Mono chromatic light of wavelength \(399 \mathrm{~nm}\) is incident from air on a water \((\mathrm{n}=1.33)\) Surface. The wavelength of refracted light is \(\mathrm{nm}\) (A) 300 (B) \(\overline{600}\) (C) 333 (D) 443

Short Answer

Expert verified
The wavelength of refracted light in water is 300 nm. The correct option is (A) 300.

Step by step solution

01

Write down the given information

The given information is: - Wavelength of incident light in air, \(\lambda_{air} = 399 \ nm\) - Refractive index of water, \(n_{water} = 1.33\)
02

Write the formula relating the wavelength in two different media

The formula relating the wavelength in two different media is given by: \(n_{1}\lambda_{1}=n_{2}\lambda_{2}\), where \(n_1\) and \(n_2\) are the refractive indices of media 1 and 2 respectively, and \(\lambda_1\) and \(\lambda_2\) are the wavelengths in media 1 and 2 respectively.
03

Plug in the given values and solve for the wavelength in water

In our case, we have: - \(n_{1} = 1\) (refractive index of air) - \(\lambda_{1} = 399 \ nm\) (wavelength of incident light in air) - \(n_{2} = 1.33\) (refractive index of water) We need to find \(\lambda_{2}\), the wavelength of refracted light in water. Using the formula: \(n_{1}\lambda_{1}=n_{2}\lambda_{2}\), \(1 \times 399 = 1.33 \times \lambda_{2} \) Now, solve for \(\lambda_{2}\): \(\lambda_{2}\) = \(\frac{399}{1.33}\) = \(300 nm\) So, the wavelength of refracted light in water is 300 nm. The correct option is (A) 300.

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Most popular questions from this chapter

A concave mirror of focal length \(20 \mathrm{~cm}\) forms an virtual image having twice the linear dimensions of the object, the position of the object will be \(\mathrm{cm}\) (A) \(7.5\) (B) \(-10\) (C) 10 (D) \(-7.5\)

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