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Chapter 16: Problem 2285

Wave light travels from an optically rarer medium to an optically denser medium its velocity decrease because of change in (A) frequency (B) wavelength (C) amplitude (D) phase

Short Answer

Expert verified
The correct answer is (B) wavelength. The wavelength of the light wave decreases when traveling from an optically rarer medium to an optically denser medium, causing the velocity to decrease as well.

Step by step solution

01

Recall Snell's Law and the Wave Equation

Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of light in the two media. \[n_1 \times \sin{\theta_1} = n_2 \times \sin{\theta_2}\] where \(n_1\) and \(n_2\) are the refractive indices of the rarer and denser media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction. The wave equation relates the velocity (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) of a wave: \[v = f \lambda\]
02

Analyze the effect on frequency

The frequency of the wave does not change when light passes from one medium to another. This is because the frequency depends on the source of light, and not on the medium through which the light is transmitted.
03

Analyze the effect on wavelength

Wavelength (\(\lambda\)) is the distance between two successive crests or troughs in a wave. As the velocity of light decreases when it enters an optically denser medium, the wavelength also decreases. This can be seen from the wave equation: \[\lambda = \frac{v}{f}\] The frequency (\(f\)) remains constant, but the velocity (\(v\)) decreases, so the wavelength (\(\lambda\)) also decreases.
04

Analyze the effect on amplitude

Amplitude is the maximum displacement of a point on a wave from its equilibrium position. The amplitude of the wave does not change when light passes from one medium to another. This is because the amplitude depends on the energy of the wave, and not on the medium through which the light is transmitted.
05

Analyze the effect on phase

Phase is the position of a point in time on a waveform cycle. Although the phase of the light wave changes as it enters a denser medium, this is not the cause of the decrease in velocity.
06

Choose the correct answer

According to our analysis, the correct answer is (B) wavelength. The wavelength of the light wave decreases when traveling from an optically rarer medium to an optically denser medium, causing the velocity to decrease as well.

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Most popular questions from this chapter

"Bhautik" runs towards a plane mirror with a speed of \(20 \mathrm{~ms}^{-1}\), what is the speed of his image ? (A) \(45 \mathrm{~ms}^{-1}\) (B) \(20 \mathrm{~ms}^{-1}\) (C) \(15 \mathrm{~ms}^{-1}\) (D) \(7.5 \mathrm{~ms}^{-1}\)

Angle of minimum deviation for a prism refractive index \(1.5\) is equal to the angle of the prism. Then the angle of prism (given, $\sin 48^{\circ} 36^{\prime}=0.75$ ) (A) \(62^{\circ}\) (B) \(82^{\circ}\) (C) \(60^{\circ}\) (D) \(41^{\circ}\)

A ray of light passes through a prism having refractive index \((\mathrm{n}=\sqrt{2})\), Suffers minimum deviation If angle of incident is double the angle of refraction within prism then angle of prism is (A) \(30^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(180^{\circ}\)

Light of wave-length \(\lambda\) is incident on a slit of width \(\mathrm{d}\). The resulting diffraction pattern is observed on a screen placed at a distance \(\mathrm{D}\). The linear width of the principal maximum is equal to the width of the slit, then \(\mathrm{D}=\) (A) \(\left(\mathrm{d}^{2} / 2 \lambda\right)\) (B) \(\left(2 \lambda^{2} / \mathrm{d}\right)\) (C) \((\mathrm{d} / \lambda)\) (D) \((2 \lambda / \mathrm{d})\)

The width of a single slit, if the first minimum is observed at an angle of \(2^{\circ}\) with a wavelength of light \(6980 \AA\) is \(\mathrm{mm}\) (A) \(0.2\) (B) \(2 \times 10^{-5}\) (C) \(2 \times 10^{5}\) (D) \(0.02\)
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