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A beam of monochromatic blue light of wavelength $4200{{A}^{{}^\circ }}$ in air travels in water of refractive index 4/3. Its wavelength in water will be:
(A) $4200{{A}^{{}^\circ }}$
(B) $5800{{A}^{{}^\circ }}$
(C) $4150{{A}^{{}^\circ }}$
(D) $3150{{A}^{{}^\circ }}$

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Last updated date: 19th Jun 2024
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Answer
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Hint: We should know that monochromatic light is light or optical radiation where the optical spectrum contains only a single optical frequency. The associated electric field strength at a certain point in space, for example, exhibits a purely sinusoidal oscillation, having a constant instantaneous frequency and a zero bandwidth. Monochromatic light is light of a single wavelength, or at least very narrow bandwidth. This makes for very sharply defined interference bands. This is pretty much essential for a good two slit result. As it is usual to use lasers for such experiments, the light will also be coherent. Normally, a laser is considered to be the best monochromatic light source. However, lasers are rather expensive and provide only single wavelengths or very small bands. An economic, broadband alternative is the combination of light source and monochromator. Based on this concept we have to solve this question.

Complete step by step answer
Let us first define refractive index as the measure of the bending of a ray of light when passing from one medium into another. When light enters a material with higher refractive index, the angle of refraction will be smaller than the angle of incidence and the light will be refracted towards the normal of the surface. The higher the refractive index, the closer to the normal direction the light will travel.
It should be known to us that refractive index values are usually determined at standard temperature. A higher temperature means the liquid becomes less dense and less viscous, causing light to travel faster in the medium. This results in a larger value for the refractive index due to a larger ratio.
We know that:
Wavelength $\lambda \propto \dfrac{1}{\mu}$
Here,
$\lambda =$wavelength of the monochromatic blue light
$\mu$is the refractive index of the monochromatic blue light
So, the ratio becomes:
$\therefore \dfrac{\lambda_{2}}{\lambda_{1}}=\dfrac{\mu_{1}}{\mu_{2}}=\dfrac{3}{4} \quad\left(\because{ }_{1} \mu_{2}=\dfrac{4}{3}\right)$
Hence, we can say that:
${{\lambda }_{2}}=\dfrac{3}{4}{{\lambda }_{1}}=\dfrac{3}{4}\times 4200=3150{{\text{A}}^{{}^\circ }}$
So, the wavelength in water will be $3150{{A}^{{}^\circ }}$.

Hence, the correct answer is option D.

Note: We should know that the wavelength is the distance between two wave crests, which is the same as the distance between two troughs. The number of waves that pass-through a given point in one second is called the frequency, measured in units of cycles per second called Hertz. As the full spectrum of visible light travels through a prism, the wavelengths separate into the colours of the rainbow because each colour is a different wavelength. Violet has the shortest wavelength, at around 380 nanometres, and red has the longest wavelength, at around 700 nanometres. Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation.