282087 A pole is vertically submerged in swimming pool, such that is gives a length of shadow 2.15 $m$ within water when sunlight is incident at an angle of ${30}^{\circ }$ with the surface of water. If swimming pool is filled to a height of $1.5\mathrm{m}$, then the height of the pole above the water surface in centimeters is $(n_w=4/3)$

Ans. (50) : Given data,
Shadow length of pole $=2.15\mathrm{m}$
Incident angle $={30}^{\circ }$
height of water in swimming pool $=1.5\mathrm{m}$
Refractive index of water $\left(n_w\right)=4/3$
Refractive index of air $\left({\mathrm{n}}_{\text{air }}\right)=1$
From snell's law,
$\begin{array}{r}1\sin {60}^{\circ }=\frac{4}{3}\sin r\\ \sin r=\frac{3\sqrt{3}}{8}\text{ this gives }\tan r=\frac{3\sqrt{3}}{\sqrt{37}}\end{array}$
From the diagram,
$\begin{array}{r}x\sqrt{3}+1.5\tan r=2.15\\ x\sqrt{3}=2.15-1.5\times \frac{3\sqrt{3}}{\sqrt{37}}\\ x\sqrt{3}=2.15-1.5\times \frac{3\sqrt{3}}{\sqrt{37}}\\ x=\frac{2.15-1.281}{\sqrt{3}}\\ x=0.50\mathrm{m}\\ x=50\mathrm{cm}\end{array}$

282088 A monochromatic light wave with wavelength ${\lambda }_1$ and frequency $v_1$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are ${45}^{\circ }$ and ${30}^{\circ }$ respectively, then the wavelength ${\lambda }_2$ and frequency $v_2$ of the refracted wave are :

282090 The refractive index of crown glass is close to $3/2$. If the speed of light in the crown glass will be close to

282087 A pole is vertically submerged in swimming pool, such that is gives a length of shadow 2.15 $m$ within water when sunlight is incident at an angle of ${30}^{\circ }$ with the surface of water. If swimming pool is filled to a height of $1.5\mathrm{m}$, then the height of the pole above the water surface in centimeters is $(n_w=4/3)$

Ans. (50) : Given data,
Shadow length of pole $=2.15\mathrm{m}$
Incident angle $={30}^{\circ }$
height of water in swimming pool $=1.5\mathrm{m}$
Refractive index of water $\left(n_w\right)=4/3$
Refractive index of air $\left({\mathrm{n}}_{\text{air }}\right)=1$
From snell's law,
$\begin{array}{r}1\sin {60}^{\circ }=\frac{4}{3}\sin r\\ \sin r=\frac{3\sqrt{3}}{8}\text{ this gives }\tan r=\frac{3\sqrt{3}}{\sqrt{37}}\end{array}$
From the diagram,
$\begin{array}{r}x\sqrt{3}+1.5\tan r=2.15\\ x\sqrt{3}=2.15-1.5\times \frac{3\sqrt{3}}{\sqrt{37}}\\ x\sqrt{3}=2.15-1.5\times \frac{3\sqrt{3}}{\sqrt{37}}\\ x=\frac{2.15-1.281}{\sqrt{3}}\\ x=0.50\mathrm{m}\\ x=50\mathrm{cm}\end{array}$

282088 A monochromatic light wave with wavelength ${\lambda }_1$ and frequency $v_1$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are ${45}^{\circ }$ and ${30}^{\circ }$ respectively, then the wavelength ${\lambda }_2$ and frequency $v_2$ of the refracted wave are :

282089 Given below are two statement:
Statement I : If the Brewster's angle for the light propagating from air to glass is ${\theta }_B$, then the Brewster's angle for the light propagating
from glass to air is $\frac{\pi }{2}-{\theta }_B$.
Statement II: The Brewster's angle for the light propagating from glass to air is ${\tan }^{-1}\left({\mu }_{\mathrm{g}}\right)$ where ${\mu }_{\mathrm{g}}$ is the refractive index of glass.
In the light of the above statement, choose the correct answer from the option given below:

282090 The refractive index of crown glass is close to $3/2$. If the speed of light in the crown glass will be close to

282087 A pole is vertically submerged in swimming pool, such that is gives a length of shadow 2.15 $m$ within water when sunlight is incident at an angle of ${30}^{\circ }$ with the surface of water. If swimming pool is filled to a height of $1.5\mathrm{m}$, then the height of the pole above the water surface in centimeters is $(n_w=4/3)$

Ans. (50) : Given data,
Shadow length of pole $=2.15\mathrm{m}$
Incident angle $={30}^{\circ }$
height of water in swimming pool $=1.5\mathrm{m}$
Refractive index of water $\left(n_w\right)=4/3$
Refractive index of air $\left({\mathrm{n}}_{\text{air }}\right)=1$
From snell's law,
$\begin{array}{r}1\sin {60}^{\circ }=\frac{4}{3}\sin r\\ \sin r=\frac{3\sqrt{3}}{8}\text{ this gives }\tan r=\frac{3\sqrt{3}}{\sqrt{37}}\end{array}$
From the diagram,
$\begin{array}{r}x\sqrt{3}+1.5\tan r=2.15\\ x\sqrt{3}=2.15-1.5\times \frac{3\sqrt{3}}{\sqrt{37}}\\ x\sqrt{3}=2.15-1.5\times \frac{3\sqrt{3}}{\sqrt{37}}\\ x=\frac{2.15-1.281}{\sqrt{3}}\\ x=0.50\mathrm{m}\\ x=50\mathrm{cm}\end{array}$

282088 A monochromatic light wave with wavelength ${\lambda }_1$ and frequency $v_1$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are ${45}^{\circ }$ and ${30}^{\circ }$ respectively, then the wavelength ${\lambda }_2$ and frequency $v_2$ of the refracted wave are :

282089 Given below are two statement:
Statement I : If the Brewster's angle for the light propagating from air to glass is ${\theta }_B$, then the Brewster's angle for the light propagating
from glass to air is $\frac{\pi }{2}-{\theta }_B$.
Statement II: The Brewster's angle for the light propagating from glass to air is ${\tan }^{-1}\left({\mu }_{\mathrm{g}}\right)$ where ${\mu }_{\mathrm{g}}$ is the refractive index of glass.
In the light of the above statement, choose the correct answer from the option given below:

282090 The refractive index of crown glass is close to $3/2$. If the speed of light in the crown glass will be close to

282087 A pole is vertically submerged in swimming pool, such that is gives a length of shadow 2.15 $m$ within water when sunlight is incident at an angle of ${30}^{\circ }$ with the surface of water. If swimming pool is filled to a height of $1.5\mathrm{m}$, then the height of the pole above the water surface in centimeters is $(n_w=4/3)$

Ans. (50) : Given data,
Shadow length of pole $=2.15\mathrm{m}$
Incident angle $={30}^{\circ }$
height of water in swimming pool $=1.5\mathrm{m}$
Refractive index of water $\left(n_w\right)=4/3$
Refractive index of air $\left({\mathrm{n}}_{\text{air }}\right)=1$
From snell's law,
$\begin{array}{r}1\sin {60}^{\circ }=\frac{4}{3}\sin r\\ \sin r=\frac{3\sqrt{3}}{8}\text{ this gives }\tan r=\frac{3\sqrt{3}}{\sqrt{37}}\end{array}$
From the diagram,
$\begin{array}{r}x\sqrt{3}+1.5\tan r=2.15\\ x\sqrt{3}=2.15-1.5\times \frac{3\sqrt{3}}{\sqrt{37}}\\ x\sqrt{3}=2.15-1.5\times \frac{3\sqrt{3}}{\sqrt{37}}\\ x=\frac{2.15-1.281}{\sqrt{3}}\\ x=0.50\mathrm{m}\\ x=50\mathrm{cm}\end{array}$

282088 A monochromatic light wave with wavelength ${\lambda }_1$ and frequency $v_1$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are ${45}^{\circ }$ and ${30}^{\circ }$ respectively, then the wavelength ${\lambda }_2$ and frequency $v_2$ of the refracted wave are :

282089 Given below are two statement:
Statement I : If the Brewster's angle for the light propagating from air to glass is ${\theta }_B$, then the Brewster's angle for the light propagating
from glass to air is $\frac{\pi }{2}-{\theta }_B$.
Statement II: The Brewster's angle for the light propagating from glass to air is ${\tan }^{-1}\left({\mu }_{\mathrm{g}}\right)$ where ${\mu }_{\mathrm{g}}$ is the refractive index of glass.
In the light of the above statement, choose the correct answer from the option given below:

282090 The refractive index of crown glass is close to $3/2$. If the speed of light in the crown glass will be close to