Answer:
The angle between the red and blue light is 1.7°.
Explanation:
Given that,
Wavelength of red = 656 nm
Wavelength of blue = 486 nm
Angle = 37°
Suppose we need to find the angle between the red and blue light as it leaves the prism
![n_{r}=1.572](https://tex.z-dn.net/?f=n_%7Br%7D%3D1.572)
![n_{b}=1.587](https://tex.z-dn.net/?f=n_%7Bb%7D%3D1.587)
We need to calculate the angle for red wavelength
Using Snell's law,
![n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}](https://tex.z-dn.net/?f=n_%7Br%7D%5Csin%5Ctheta_%7Bi%7D%3Dn_%7Ba%7D%5Csin%5Ctheta_%7Br%7D)
Put the value into the formula
![1.572\sin37=1\times\sin\theta_{r}](https://tex.z-dn.net/?f=1.572%5Csin37%3D1%5Ctimes%5Csin%5Ctheta_%7Br%7D)
![\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})](https://tex.z-dn.net/?f=%5Ctheta_%7Br%7D%3D%5Csin%5E%7B-1%7D%28%5Cdfrac%7B1.572%5Csin37%7D%7B1%7D%29)
![\theta_{r}=71.0^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_%7Br%7D%3D71.0%5E%7B%5Ccirc%7D)
We need to calculate the angle for blue wavelength
Using Snell's law,
![n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}](https://tex.z-dn.net/?f=n_%7Bb%7D%5Csin%5Ctheta_%7Bi%7D%3Dn_%7Ba%7D%5Csin%5Ctheta_%7Bb%7D)
Put the value into the formula
![1.587\sin37=1\times\sin\theta_{b}](https://tex.z-dn.net/?f=1.587%5Csin37%3D1%5Ctimes%5Csin%5Ctheta_%7Bb%7D)
![\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})](https://tex.z-dn.net/?f=%5Ctheta_%7Bb%7D%3D%5Csin%5E%7B-1%7D%28%5Cdfrac%7B1.587%5Csin37%7D%7B1%7D%29)
![\theta_{b}=72.7^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_%7Bb%7D%3D72.7%5E%7B%5Ccirc%7D)
We need to calculate the angle between the red and blue light
Using formula of angle
![\Delta \theta=\theta_{b}-\theta_{r}](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D%5Ctheta_%7Bb%7D-%5Ctheta_%7Br%7D)
Put the value into the formula
![\Delta \theta=72.7-71.0](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D72.7-71.0)
![\Delta \theta=1.7^{\circ}](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D1.7%5E%7B%5Ccirc%7D)
Hence, The angle between the red and blue light is 1.7°.
There is no equation here
Answer:0.114 C
Explanation:
Given
Total 4.7 C is distributed in two spheres
Let
and
be the charges such that
![q_1+q_2=4.7](https://tex.z-dn.net/?f=q_1%2Bq_2%3D4.7)
and Force between charge particles is given by
![F=\frac{kq_1q_2}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bkq_1q_2%7D%7Br%5E2%7D)
![4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}](https://tex.z-dn.net/?f=4.7%5Ctimes%2010%5E11%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%20q_1%5Ccdot%20q_2%7D%7B0.1%5E2%7D)
![q_1\cdot q_2=0.522](https://tex.z-dn.net/?f=q_1%5Ccdot%20q_2%3D0.522)
put the value of ![q_1](https://tex.z-dn.net/?f=q_1)
![q_2\left ( 4.7-q_2\right )=0.522](https://tex.z-dn.net/?f=q_2%5Cleft%20%28%204.7-q_2%5Cright%20%29%3D0.522)
![q_2^2-4.7q_2+0.522=0](https://tex.z-dn.net/?f=q_2%5E2-4.7q_2%2B0.522%3D0)
![q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}](https://tex.z-dn.net/?f=q_2%3D%5Cfrac%7B4.7%5Cpm%20%5Csqrt%7B4.7%5E2-4%5Ctimes%201%5Ctimes%200.522%7D%7D%7B2%7D)
![q_2=0.114 C](https://tex.z-dn.net/?f=q_2%3D0.114%20C)
thus ![q_1=4.586 C](https://tex.z-dn.net/?f=q_1%3D4.586%20C)
As the distance between two charged objects increases, the strength of the electrical force between the objects <em>decreases</em>.