The refractive index for glycerine is

, while for air it is

.
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:

where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is
Explanation:
From the question we are told that
The focal length is 
This negative sign shows the the microscope is diverging light
The angular magnification is 
The distance between the objective and the eyepieces lenses is 
Generally the magnification is mathematically represented as
![m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]](https://tex.z-dn.net/?f=m%20%20%3D%20%20%5B%5Cfrac%7BZ%20-%20f_e%20%7D%7Bf_e%7D%5D%20%5B%5Cfrac%7B0.25%7D%7Bf_0%7D%20%5D)
Where
is the eyepiece focal length of the microscope
Now making
the subject of the formula
![f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }](https://tex.z-dn.net/?f=f_e%20%20%3D%20%5Cfrac%7BZ%7D%7B1%20-%20%5B%5Cfrac%7BM%20%20%2A%20%20f_o%20%7D%7B0.25%7D%5D%20%7D)
substituting values
![f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }](https://tex.z-dn.net/?f=f_e%20%20%3D%20%5Cfrac%7B%200.19%20%7D%7B1%20-%20%5B%5Cfrac%7B300%20%20%2A%20%20-0.0055%20%7D%7B0.25%7D%5D%20%7D)
<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>
When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:

To maintain
constant, if Q increases V also increases.
On the other hand, the potential energy
can be expressed as:

In conclusion, as Q increases the potential energy also increases.
Explanation:
first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s
now using the 3rd law of motion
v^2=u^2+2as
0=9+2a2
a= -9/4m/s^2
now force=mass×accelration
=2kg×(-9/4)m/s^2
=4.5 N
4.5 newton force applied on the book!
✌️:)