Answer:
a
b
c
Explanation:
From the question we are told that
The angle of incidence is 
The refractive index of water is 
Generally Snell's law is mathematically represented as
Here 
is the refractive index of air with value 
is the angle of refraction
So
![\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]](https://tex.z-dn.net/?f=%5Ctheta%20_2%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_1%20%2A%20sin%28%5Ctheta%20_1%29%7D%7Bn_2%7D%20%5D)
=> ![\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]](https://tex.z-dn.net/?f=%5Ctheta%20_2%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B1.3%20%2A%20sin%2810%29%7D%7B1%7D%20%5D)
=>
Given that the angle should not be greater than 
then the angle of incidence will be
![\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]](https://tex.z-dn.net/?f=%5Ctheta%20_1%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_2%20%2A%20sin%28%5Ctheta%20_2%29%7D%7Bn_1%7D%20%5D)
=> ![\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]](https://tex.z-dn.net/?f=%5Ctheta%20_1%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B1%20%2A%20sin%2845%29%7D%7B1.3%7D%20%5D)
=>
Generally for critical angle is mathematically represented as
![\theta_c = sin^{-1}[\frac{n_2}{n_1} ]](https://tex.z-dn.net/?f=%5Ctheta_c%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_2%7D%7Bn_1%7D%20%5D)
=> ![\theta_c = sin^{-1}[\frac{1}{1.3} ]](https://tex.z-dn.net/?f=%5Ctheta_c%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7B1.3%7D%20%5D)
=>
Answer:
100 ÷ 9.58 = 10.44 (approximate answer)
The first step is the fusion of hydrogen into Deuterium.
<span>11.823 cm
There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall.
The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm.
Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So
82 + 1/21 * 2 = 82 + 2/21 = 82.0952381
Now we have the following dimensions with a circle replacing the ball in the original problem.
Distance from wall to effective circle = 82.0952381 cm
Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm
Effective diameter of circle = 3.995462279 cm
And because the geometry makes similar triangles, the following ratio applies.
3.995462279/41.9047619 = X/124
Now solve for X
3.995462279/41.9047619 = X/124
124*3.995462279/41.9047619 = X
495.4373226/41.9047619 = X
11.82293611 = X
The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
