Question

When light of wavelength \[5000\mathop {\text{A}}\limits^ \circ \] in vacuum passes through water, the wavelength and frequency, if the refractive index of water is\[\dfrac{4}{3}\], are:
$\left( {\text{A}} \right){\text{ }}3750\mathop {\text{A}}\limits^ \circ ,6{\text{x1}}{{\text{0}}^{14}}{\text{Hz}}$
$\left( {\text{B}} \right){\text{ }}4000\mathop {\text{A}}\limits^ \circ ,5{\text{x1}}{{\text{0}}^{14}}{\text{Hz}}$
$\left( {\text{C}} \right){\text{ }}5000\mathop {\text{A}}\limits^ \circ ,4{\text{x1}}{{\text{0}}^{14}}{\text{Hz}}$
$\left( {\text{D}} \right){\text{ }}3500\mathop {\text{A}}\limits^ \circ ,5{\text{x1}}{{\text{0}}^{14}}{\text{Hz}}$

Answer 1

Hint:Skew angle: It is defined as the angle between perpendicular to the alignment of the bridge and the centerline of the pier.
Refraction of light: When light travels from one medium to another medium due to the change in the density of the medium the path of the light changes.
Refractive index: It is known as the ratio of the light speed in a vacuum and the light speed in a given medium.

Laws of refraction:

First law: The incident ray, the refracted ray, and normal to the interface at the point of incidence all lie in the same plane.

Second law: This law in optical physics defines that the ratio of the sine of the angle of incidence and the sine of the angle of refraction is constant for a given pair of media.

Formula used:
${\lambda _{\text{m}}} = \dfrac{{{\lambda _0}}}{{\text{n}}}$, ${\lambda _{\text{m}}} = $ wavelength of light in a medium, ${\lambda _{\text{0}}} = $wavelength of light in vacuum, n= refractive index of the medium
Frequency $(\nu ) = \dfrac{{\text{c}}}{{{\lambda _0}}}$, here, c= velocity of the light wave.

Complete step-by-step solution: Given details:
${\lambda _0} = 5000\mathop {\text{A}}\limits^ \circ = 5000{\text{x1}}{{\text{0}}^{ - 10}}\mathop {\text{A}}\limits^ \circ = 5{\text{x1}}{{\text{0}}^{ - 7}}\mathop {\text{A}}\limits^ \circ $
 Therefore, by using the equation we can write that, ${\lambda _{\text{m}}} = \dfrac{{{\lambda _0}}}{{\text{n}}}$
${\lambda _{\text{m}}} = \dfrac{{5{\text{x1}}{{\text{0}}^{ - 7}}}}{{(4/3)}}$$ = 3750\mathop {\text{A}}\limits^ \circ $
Frequency $(\nu ) = \dfrac{{\text{c}}}{{{\lambda _0}}}$
$\nu = \dfrac{{3{\text{x1}}{{\text{0}}^8}}}{{5{\text{x1}}{{\text{0}}^{ - 7}}}}{\text{Hz}}$$ = 6{\text{x1}}{{\text{0}}^{14}}{\text{Hz}}$

Hence the correct option is \[\left( A \right)\].

Note:When light travels from a denser medium to the rarer medium the refraction of light occurs away from the normal.
On seeing the coin placed at the bottom of the container from the skew point, due to the change in the density of the medium the coin will look like it is floating in the water.
The apparent depth of the coin will rise and the coin will be seen as it is floating on the water from the skew point.